Math in Focus

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Review Test Answer Key

Concepts and Skills

Match each of the solid figures to its net.

Answer:
The net form of Figure 1 is Figure c.
The net form of Figure 2 is Figure a.
The net form of Figure 3 is Figure b.
The net form of Figure 4 is Figure e.
The net form of Figure 5 is Figure d.

Find the surface area of each solid.

Question 6.

Answer:
Given that the length of the cube is 8 in.
The given figure has 6 faces.
Area of 1 face will be 8×8 = 64 sq.in
Surface area of the given solid = 6×base area
The surface area of the given solid will be 6×64 = 384 sq.in

Question 7.

Answer:
Given solid has four triangular faces and one square base.
Area of base will be 8×8 = 64 sq.m
Area of triangle = \(\frac{1}{2}\) × base × width
Area of triangular face will be \(\frac{1}{2}\) × 8 × 10 = 40 sq.m
There are 4 such triangular faces, 4×40 = 160 sq.m
The surface area of the given figure will be 64+160 = 224 sq.m

Find the volume of each prism.

Question 8.

Answer:
Given cube has length of 7 cm.
Area of base = 7×7 = 49 sq.cm
Volume of the prism = base area × height
Volume of the prism will be 49×7 = 343 cu.cm

Question 9.

Answer:
Area of triangular base = \(\frac{1}{2}\) ×base×height
Area of the base will be \(\frac{1}{2}\) ×6×5 = 15 sq.ft
Volume of the given prism will be 3×15 = 45 cu.ft

Solve.

Question 10.
The solid below is made up of cubes, each of which has an edge length of 3 inches.
a) What is the volume of one cube?
Answer:
Given that the length of the cube is 3 inches.
Area of one cube will be = 3×3 = 9 sq.in
Volume of the cube= base area × height
Volume of one cube will be = 9×3 = 27 cu.in

b) What is the volume of the solid figure?

Answer:
The below solid figure has 10 identical cubes.
Therefore, the volume of the solid figure will be 10×27 = 270 cu.in

Problem Solving

Solve.

Question 11.
A fish tank is 50 centimeters long, 30 centimeters wide, and 40 centimeters high. It contains water up to a height of 28 centimeters. How many more cubic centimeters of water are needed to fill the tank to a height of 35 centimeters?
Answer:
Given that a fish tank is 50 centimeters long, 30 centimeters wide, and 40 centimeters high.
Volume of the tank = length × width × height
Volume of water filled up to a height of 28 centimeters is 50×30×28 = 42000 cu.cm
Volume of water to fill up to 35 cm will be 50×30×35 = 52500 cu.cm
More cubic centimeters of water are needed to fill the tank to a height of 35 centimeters will be 52500-42000 = 10500 cu.cm

Question 12.
Find the surface area of a square pyramid given that its base area is 196 square inches and the height of each of its triangular faces is 16 inches.

Answer:
Given that the base area of the given square pyramid is 196 sq.in.
Area of square = 196
196 can also be written as 14 × 14
length × length = 196
Length of one side of the square will be 14 cm.
Area of trinagular face will be \(\frac{1}{2}\) × 14 × 16
= 7 × 16
= 112 cu.in
Surface area of the given solid is 4×112 + 196
= 448 + 196
= 644 sq.in

Question 13.
The volume of a rectangular prism is 441 cubic feet. It has a square base with edges that are 7 feet long.
a) Find the height of the prism.
Answer:
Given that the volume of a rectangular prism is 441 cubic feet and it has a square base with edges that are 7 feet long.
Volume of the prism = Area of the base × height
Area of the base = 7×7 = 49 sq.ft
441 = 49 × height
height = 441÷49
height = 9 ft.
Therefore, the height of the prism is 9 ft.

b) Find the surface area of the prism.

Answer:
Area of the square base = length×length = 49 sq.ft
Area of the rectangular base = length×width= 7×9 = 63 sq.ft
There are 2 square bases and 4 rectangular faces.
Therefore, the surface area of the prism will be 2×49 + 4×63
= 98 + 252
= 350 sq.ft

Question 14.
The volume of a rectangular tank with a square base is 63,908 cubic centimeters. Its height is 64 centimeters. Find the length of an edge of one of the square bases. Round your answer to the nearest tenth of a centimeter.

Answer:
Given that the volume of a rectangular tank with a square base is 63,908 cubic centimeters and it is 64 cm high.
Volume of a rectangular prism = Area of base × height
63,908 = Area of base × 64
Area of base will be 63908 ÷ 64 = 998.56 sq.cm
Area of the base when rounded to nearest ten will be 1000.
Area of base = length × length
1000 = length × length
10 × 10 = length × length
Length of an edge of one of the square bases will be 10 cm.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.4 Real-World Problem: Surface Area and Volume detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.4 Answer Key Real-World Problem: Surface Area and Volume

Math in Focus Grade 6 Chapter 12 Lesson 12.4 Guided Practice Answer Key

Complete.

Question 1.
Find the volume of water needed to fill three fourths of the aquarium.

Height of water needed = \(\frac{3}{4}\) •
= in.
Volume of water needed = • •
= in.³
The aquarium needs to have cubic inches of water added to it to be \(\frac{3}{4}\) full.
Answer:
Volume = length × width × height
Volume of water needed = 25×12×14
= 300×14
= 4200 cu.in
Height of water needed = \(\frac{3}{4}\) × 4200
= 3 × 150
= 3150 cu.in
The aquarium needs to have 3150 cubic inches of water added to it to be \(\frac{3}{4}\) full.

Question 2.
A metal bar has bases that are parallelograms.

a) Find the volume of the metal bar.
Area of parallelogram
= base of parallelogram • height of parallelogram
= •
= cm²
Volume of metal bar = base of prism • height of prism
= •
= cm³
The volume of the metal bar is cubic centimeters.
Answer:
Area of parallelogram
= base of parallelogram × height of parallelogram
= 8 × 4
= 32 cm²
Volume of metal bar = base of prism × height of prism
= 32 × 24
= 768  cm³
The volume of the metal bar is 480 cubic centimeters.

b) Find the surface area of the metal bar.
Surface area of metal bar
= perimeter of base • height + total area of 2 bases

The surface area of the metal bar is square centimeters.
Answer:
Given parallelogram is 5cm high, 24cm long and 8cm wide.
The perpendicular height of the base is 4cm.
Surface area of metal bar
= perimeter of base • height + total area of 2 bases
= (24+8+8+24) × 5 + (20+20)
= (64×5) + 40
= 320 + 40
= 360 sq.cm

Question 3.
A candle is a square prism. The candle is 15 centimeters high, and its volume is 960 cubic centimeters.

a) Find the length of each side of the square base.
V = Bh
= B •
÷ = • B ÷
= B
Length of each side of base
=
= cm
The length of each side of the square base is centimeters.
Answer:
Given that the square prism is 15cm high and its volume is 960 cubic centimeters.
V = Bh
960 = B × 15
960 ÷ 15 = 15 × B ÷ 15
64 = B
Base is side × side = 8 × 8
The length of each side of the square base is 8 centimeters.

b) Find the surface area of the candle.
Surface area of candle
= perimeter of base • height + area of two bases

The surface area of the candle is square centimeters.
Answer:
= perimeter of base × height + area of two bases
= (8 + 8 +8 + 8) × 15 + (2×64)
= 32 × 15 + 128
= 480 + 128
= 608 sq.cm

Question 4.
A storage chest is a prism with bases that are pentagons. The diagram shows some of the dimensions of the storage chest. The volume of the storage chest is 855 cubic inches.

a) Find the height AS of the prism. Round your answer to the nearest hundredth.
Area of pentagonal base = area of trapezoid + area of rectangle
= \(\frac{1}{2}\) • • ( + ) + •
= +
= in.²

Answer:
Area of pentagonal base = area of trapezoid + area of rectangle
= \(\frac{1}{2}\) × 3× (3+7) + 7× 9
= 3×5 + 63
= 15+63
= 78 sq.in

V = Bh
= • h
÷ = • h ÷
= ≈ h
The height of the prism is approximately inches.
Answer:
855 = 78×h
855 ÷ 78 = 78 × h ÷ 78
10.96 = h
Height of the prism will be 10.96 in.

b) Find the surface area of the prism. Round your answer to the nearest hundredth.
Surface area of prism
= perimeter of base • height + area of two bases

The surface area of the prism is approximately square inches.
Answer:
Area of two bases = 78 + 78 = 156 sq.in
= perimeter of base • height + area of two bases
= (5+3+3+9+7)×10.96 + 156
= 27×10.96 + 156
= 295.92 + 156
= 451.92 sq.in
The surface area of the prism is approximately 500 sq.inf

Math in Focus Course 1B Practice 12.4 Answer Key

Solve.

Question 1.
Savannah has a water bottle that is a rectangular prism. The bottle measures 7 centimeters by 5 centimeters by 18 centimeters and she filled it completely with water. Then, she drank \(\frac{1}{4}\) of the volume of water in her water bottle. How many cubic centimeters of water were left in the water bottle?

Answer:
Given that the bottle measures 7 centimeters by 5 centimeters by 18 centimeters and she filled it completely with water.
Therefore, the volume of the rectangular prism = length×width×height
= 7×5×18
= 630 cu.cm
Volume of water in the bottle is 630 cu.cm
Out of which she drank \(\frac{1}{4}\), so the left over water will be \(\frac{1}{4}\) of total volume
= \(\frac{1}{4}\) × 630
= 157.5 cu.cm
Therefore, 157.5 cu.cm of water will be left in the water bottle.

Question 2.
A rectangular prism has a square base with edges measuring 8 inches each. Its volume is 768 cubic inches.
a) Find the height of the prism.
Answer:
Given that a rectangular prism has a square base with edges measuring 8 inches each and its volume is 768 cubic inches.
Volume of rectangular prism = 768
Area of base × height = 768
8×8× height = 768
height = 768÷64
Height of the prism is 12 in

b) Find the surface area of the prism.
Answer:
Surface area of the base = 8×8 = 64sq.in
Total area of 2 bases = 2×64 = 128 sq.in
Surface area of the prism = perimeter of base × height + total area of 2 bases
= 4×8×12 + 128
= 144 +128
= 272 sq.in

Question 3.
A triangular prism has the measurements shown.
a) Find the volume of the prism.
Answer:
Area of triangular base = \(\frac{1}{2}\)×base×height
Area of triangular base = \(\frac{1}{2}\)×19.6×5
= 49 sq.ft
`Volume of triangular prism = 49×16 = 784 cu.ft

b) Find the surface area of the prism.

Answer:
Area of triangular base = \(\frac{1}{2}\)×base×height
Area of triangular base = \(\frac{1}{2}\)×19.6×5
= 49 sq.ft
Area of two triangular faces will be 2×49 = 98 sq.ft
Area of rectangular face = length×width
Area of first rectangular face = 16×10 = 160 sq.ft
Area of second rectangular face = 14×16 = 192 sq.ft
Area of third rectangular face = 16×19.6 = 313.6 sq.ft
Surface area of prism = 98+160+192+313.6
= 763.6 sq.ft

Question 4.
The volume of Box A is \(\frac{2}{5}\) the volume of Box B. What is the height of Box A if it has a base area of 32 square centimeters?

Answer:
Box B:
Box B is 16cm wide, 8cm long and 10cm high.
Volume of Box B will be = 16×8×10 = 1280 cu.cm
Given that volume of Box A is \(\frac{2}{5}\) the volume of Box B.
Therefore, volume of Box A = \(\frac{2}{5}\)×1280
= 2×256
= 512
Volume of Box A = Area of Base × height
512 = 32 × height
Height = 16 cm.
Therefore height of Box A will be 16cm.

Question 5.
The ratio of the length to the width to the height of an open rectangular tank is 10 : 5 : 8. The height of the tank is 18 feet longer than the width.
a) Find the volume of the tank.
Answer:
Let the width of the tank be ‘a’ ft.
Height of the tank is 18 feet longer than the width, therefore, it will be 18+a
If height equals to 8 parts, width equals 5 parts, then 8 – 5 parts = 18 ft
3 parts = 18 ft.
1 part = 6ft
Therefore, now the length will be 10×6=60 ft, width will be 5×6=30ft and height will be 8×6=48ft.
Volume of the tank will be 60×30×48=86400 cu.ft

b) Find the surface area of the open tank.
Answer:
Since the rectangular tank resembles rectangular prism.
Surface area of rectangular prism = 2×(length×width + width×height + height×length)
The surface area of the open tank = 2×(60×30 + 30×48 + 48×60)
= 12240 sq.ft

Question 6.
Janice is making a gift box. The gift box is a prism with bases that are regular hexagons, and has the dimensions shown in the diagram.

a) Find the height PQ of the prism.
Answer:
Given that the volume of prism is 2835 cu.cm
Volume of prism = Area of hexagonal base × height
6 equilateral triangles form a regular hexagon.
Therefore, Area of hexagonal base = 6×(\(\frac{1}{2}\)×7×6)
Area of hexagonal base = 6×(\(\frac{1}{2}\)×7×6)
= 6×(7×3)
= 6×21
= 126 sq.cm
2835 = 126 × height
Height = 2835÷126
height = 22.5cm
The height PQ of the prism will be 22.5cm

b) Find the surface area of the prism.
Answer:
Surface area of prism will be = Perimeter × height + Area of base
= 7×6 ×22.5 + 126 + 126
= 42×22.5 + 252
= 945 + 252
= 1197 sq.cm
The surface area of the prism will be 1197 sq.cm

Question 7.
Container A was filled with water to the brim. Then, some of the water was poured into an empty Container B until the height of the water in both containers became the same. Find tT.

Answer:
Container A:
Container A is a rectangular prism of 40cm high, 25cm long and 30 cm wide.
Volume of the rectangular prism is = length × width × height
Volume of water in the container will be 40×25×30 = 30000 cu.cm
The volumes of two containers will be same.
Container B:
Therefore the volume of Container B will be 30000 cu.cm
Given that the container B is 18cm long and 25cm wide.
Volume of the rectangular prism is = length × width × height
30000 = length × width × height
30000 = 18 × 25× height
Height = 30000÷450
Height = 66.6cm.
The new height of the water in both containers will be 66.6cm

Brain @ Work

Question 1.
The volume of a cube is 1000 cubic inches. If each of the edges is doubled in length, what will be the volume of the cube?
Answer:
Given that the volume of a cube is 1000 cubic inches.
Volume of a cube = Area of base × height
1000 =length × width × height
1000 = 10 × 10 × 10
Therefore, we can say length of the cube will be 10 in.
If the length of edge is doubled, then it will be 2×10 = 20 in.
Volume of the new cube will be 20×20×20 = 8000 cu.in

Question 2.
The volume of a cube is x cubic feet and its surface area is x square feet, where x represents the same number. Find the length of each edge of the cube.
Answer:
Given that the volume of a cube is x cubic feet and its surface area is x square feet, where x represents the same number.
Volume of the cube = Area of base × height
Surface area will be 6×Area of base
Given surface area is x sq.ft
x = 6 × Area of base
Area of base = \(\frac{x}{6}\)
Volume of the cube = \(\frac{x}{6}\) × height
x = \(\frac{x}{6}\) × height
Height = \(\frac{6x}{x}\)
Height will be 6 ft.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.3 Volume of Prisms detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.3 Answer Key Volume of Prisms

Math in Focus Grade 6 Chapter 12 Lesson 12.3 Guided Practice Answer Key

Find the volume of each rectangular prism.

Question 1.

Length = 5\(\frac{1}{4}\) in.
Width = 6 in.
Height = 12 in.
V = lwh
= • •
= in.³
Answer:
V = lwh
Length = 5\(\frac{1}{4}\) in.
(5×\(\frac{4}{4}\)+\(\frac{1}{4}\)) = \(\frac{21}{4}\)
V = \(\frac{21}{4}\) × 6 × 12 = 378 cu.in

Question 2.

Length = 8 cm
Width = 7.2 cm
Height = 3 cm
V = lwh
= • •
= cm³
Answer:
V = 8 × 7.2 × 3
172.8 cu.cm

Question 3.

Length = 4 ft
Width = 3 ft
Height = 8 \(\frac{1}{3}\) ft
Volume =
= • •
= ft³
Answer:
V = lwh
Height = 8 \(\frac{1}{3}\) ft
(8 × \(\frac{3}{3}\))+\(\frac{1}{3}\) = \(\frac{25}{3}\)
Volume = 4×3×\(\frac{25}{3}\) = 100 cu.ft

Tell whether slices parallel to each given slice will form uniform cross sections. If not, explain why not.

Question 4.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is larger than the area’s of the cube’s face.

Question 5.

Answer:
In the above given figure, the slice parallel to each given slice will form uniform cross-sections.

Question 6.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the rectangular’s face.

Find the volume of each prism.

Question 7.

Length = 6 cm
Width = 5.5 cm
Height = 9 cm
Area of base = •
= cm²
Volume of prism = •
= cm²
The volume of the prism is cubic centimeters.
Answer:
Length = 6 cm
Width = 5.5 cm
Height = 9 cm
Area of base = 6×5.5 = 33 sq.cm
Volume of prism = Base area of prism × height
Volume of prism = 33×9 = 297 cu.cm
The volume of the prism is 297 cubic centimeters.

Question 8.

Base of triangle = 10 in.
Height of triangle = 3\(\frac{1}{2}\) in.
Height of prism = 14 in.
Area of base = \(\frac{1}{2}\) • •
= in.²
Volume of prism = •
= in.³
The volume of the prism is cubic inches.
Answer:
Base of triangle = 10 in.
Height of triangle = 3\(\frac{1}{2}\) in.
(3×\(\frac{2}{2}\))+\(\frac{1}{2}\) = \(\frac{7}{2}\)
Height of prism = 14 in.
Area of triangular base = \(\frac{1}{2}\) × base × height
Area of base = \(\frac{1}{2}\) × 10 × \(\frac{7}{2}\) = \(\frac{35}{2}\) sq.in
Volume of prism = Base area of prism × height
Volume of prism = \(\frac{35}{2}\) × 14 = 245 cu.in

Question 9.

Length of shorter base of trapezoid = 4 ft
Length of longer base of trapezoid = 10 ft
Height of trapezoid = 2 ft
Height of prism = 12 ft
Area of base = \(\frac{1}{2}\) • ( + )
= \(\frac{1}{2}\) • •
= ft²
Volume of prism = •
= ft³
The volume of the prism is cubic feet.
Answer:
The formula to find the base area = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of base = \(\frac{1}{2}\) × 2 × (4+10)
= \(\frac{1}{2}\) × 2 × 14
= 14 sq.ft
Volume of prism = Base area of prism × height
Volume of prism = 14×12
= 168 cu.ft

Hands-On Activity

DETERMINING THE RELATIONSHIP BETWEEN VOLUME AND SURFACE AREA OF PRISMS

Materials:

• 27 unit cubes.

Work in pairs.
Step 1: Build the cube and the rectangular prism using unit cubes.

Step 2: Find the volume of the cube. Find the volume of the rectangular prism. What can you say about the volumes of the cube and the rectangular prism?
Answer:
Volume of cube = lwh
Volume of the cube = 2×2×2 = 8 cu.cm
Volume of the rectangular prism = 2×4×1 = 8 cu.cm
The volumes of the cube and the rectangular prism are same.

Step 3: Find the surface area of the cube. Draw its net if it helps you. Find the surface area of the rectangular prism. Draw its net if it helps you. What can you say about the surface areas of the cube and the rectangular prism?
Answer:
Area of one square will be 2×2 = 4 sq.cm
Volume of cuboid = 6 × Area of base
There are 6 faces for a cube, therefore the surface area will be 6×4 = 24 sq.cm
The surface areas of the rectangular prism will be 2(2×4 + 2×1 + 1×4) = 28 sq.cm
The surface areas of the cube and the rectangular prism are different.

Step 4: Now build these rectangular prisms using unit cubes.

Step 5: Find the volume of the cube. Find the volume of the rectangular prism. What can you say about their volumes?
Answer:
Volume of the cube = 3×3×3 = 27 cu.cm
Volume of the rectangular prism = 3×9×1 = 27 cu.cm
The volumes of the cube and the rectangular prism are same.

Step 6: Find the surface area of the cube. Find the surface area of the . rectangular prism. Draw their nets if it helps you. What can you say about their surface areas?
Area of one square will be 3×3 = 9 sq.cm
There are 6 faces for a cube, therefore the surface area will be 6×9 = 54 sq.cm
The surface areas of the rectangular prism will be 2(3×9 + 9×1 + 1×3) = 78 sq.cm
The surface areas of the cube and the rectangular prism are different.

Math Journal Based on the activity, what can you conclude about prisms with the same volume? Discuss with your partner and explain your thinking.
Answer:
Two prisms of different measurements might have the same volume, they might not have the same surface area.
For example: A rectangular prism with side lengths of 1 cm, 2 cm, and 2 cm has a volume of 4 cu cm and a surface area of 16 sq cm. A rectangular prism with side lengths of 1 cm, 1 cm, and 4 cm has the same volume but a surface area of 18 sq cm

Math in Focus Course 1B Practice 12.3 Answer Key

Solve.

Question 1.
A cube has edges measuring 9 inches each. Find the volume of the cube.
Answer:
Given that: A cube has edges measuring 9 inches each.
The base area will be equal to 9×9=81 sq.in
Volume of cube = Area of base × height
Volume of the cube will be 81×9=729 cu.in

Question 2.
A cube has edges measuring 6.5 centimeters each. Find the volume of the cube.
Answer:
Given that: A cube has edges measuring 6.5 cm each.
The base area will be equal to 6.5×6.5=42.25 sq.cm
Volume of cube = Area of base × height
Volume of the cube will be 81×9=274.625 cu.cm

Question 3.
A storage container is shaped like a rectangular prism. The container is 20 feet long, 10 feet wide, and 5\(\frac{1}{2}\) feet high. Find the volume of the storage container.
Answer:
Given that: A rectangular prism is 20 feet long, 10 feet wide, and 5\(\frac{1}{2}\) feet high.
Area of base will be 20×10=200 sq.ft
Volume of prism = Area of base × height
Volume of the storage container will be 200×5\(\frac{1}{2}\)
= 200×(5×\(\frac{2}{2}\) + \(\frac{1}{2}\))
= 200×(\(\frac{10}{2}\) + \(\frac{1}{2}\))
= 200×\(\frac{11}{2}\)
= 100×11
= 1100 cu.ft

Question 4.
Find the volume of the peppermint tea box on the right.

Answer:
Given that: The peppermint tea box is 12.6 cm long, 6.7 cm wide, and 7.8 cm high.
Area of  base will be 12.6×6.7 = 84.42 sq.cm
Volume of prism = Area of base × height
Volume of the box will be 7.8×84.42 = 658.476 cu.cm

Question 5.
The solid below is made of idential cubes. Each cube has an edge length of 2 inches. Find the volume of the solid.

Answer:
Given that: Each cube has an edge length of 2 inches.
Let us first find the volume of a single cube
Area of the base of a single cube will be 2×2=4 sq.in
Volume of cube = Area of base × height
Volume of a single cube will be  4×2=8 cu.in
The given solid is made of idential cubes. There are 9 such cubes.
Therefore, the volume of all the 9 identical cubes or the given solid figure will be 9×8=72 cu.in

Find the volume of the triangular prism.

Question 6.

Answer:
Given that: A triangular prism is 15ft long, 10ft wide and 6ft high.
Area of the triangular prism = \(\frac{1}{2}\) × base × height
Area of the triangular prism will be \(\frac{1}{2}\) × 10 × 6
= 10 × 3
= 30 sq.ft
Volume of prism = Area of base × height
Volume of the given triangular prism will be 30×15=450 cu.ft

Question 7.

Answer:
Given that: A triangular prism is 12cm long, 6.7cm wide and 3cm high.
Area of the triangular prism = \(\frac{1}{2}\) × base × height
Area of the triangular prism will be \(\frac{1}{2}\) × 6.7 × 3
= \(\frac{1}{2}\) × 20.1
= 10.05 sq.cm
Volume of prism = Area of base × height
Volume of the given triangular prism will be 10.05×12=120.6 cu.cm

Tell whether slices parallel to each given slice will form uniform cross-sections. If not, explain why not.

Question 8.

Answer:
In the above given figure, the slice parallel to each given slice will form uniform cross-sections.

Question 9.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the circular face.

Question 10.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the rectangular’s face.

Copy the solid. Draw a slice that has the same cross section as the bases In each prism.

Question 11.

Answer:

The bases for the above given figure is square.
Therefore, a slice that has the same cross section is drawn for the given prism.

Question 12.

Answer:

The bases for the above given figure is hexagon.
Therefore, a slice that has the same cross section is drawn for the given prism.

Question 13.

Answer:

The bases for the above given figure is triangle.
Therefore, a slice that has the same cross section is drawn for the given prism.

Solve.

Question 14.
The bases of the prism shown are trapezoids. Find the volume of the prism.

Answer:
Length of shorter base of trapezoid = 2 m
Length of longer base of trapezoid = 6 m
Height of trapezoid = 2 m
Height of prism = 10 m
The area of the given figure = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of base = \(\frac{1}{2}\) × 2 × (2+6)
= \(\frac{1}{2}\) × 2 × 8
= 8 sq.m
Volume of prism = base area × height
The volume of prism = 10×8
= 80 cu.m

Question 15.
A cube has a volume of 125 cubic inches. Find the length of its edge.
Answer:
Given that: A cube has a volume of 125 cu.in
Let us assume the length of a cube be ‘a’ unit.
As cube has all equal sides, the base area will be a×a = a² sq.units
Now, the volume will be a²×a = a³ cubic units.
Therefore we can say that the volume of a cube is the length of an edge taken to the third power.
a³ = 125 and 125 in cube can be written as 5³
a³ = 5³
a = 5
Therefore, the length of its edge will be 5 inches.

Question 16.
The volume of a triangular prism is 400 cubic centimeters. Two of its dimensions are given in the diagram. Find the height of a triangular base.

Answer:
Given that: The volume of a triangular prism is 400 cubic centimeters.
The three rectangular faces measures 10cm long and 8cm wide.

The area of three rectangular faces will be 3×10×8 = 240 sq.cm.
Inorder to get the areas of the triangular bases, we need to subtract the area of the rectangular faces from the total volume.
Area of two triangular bases will be 400-240 = 160 sq.cm
Area of one triangular base will be 160÷2 = 80 sq.cm
From the given figure, the base length of the triangle is 10 cm and the height is ‘h’ cm.
Area of triangular base = \(\frac{1}{2}\) × base × height
Area of the triangle is \(\frac{1}{2}\)×10×h = 80
5×h = 80
h = 80÷5
h = 16 cm.

Question 17.
A cross-section of the triangular prism shown below is parallel to a base. The area of the cross-section is 24 square feet. The ratio of DM to MA is 3 : 5 and the length of \(\overline{F O}\) is 6 feet. Find the volume of the triangular prism.

Answer:
Given that the area of cross-section is 24 sq.ft and ratio of DM to MA is 3 : 5.
The length of FO is 6 ft.
By observing the FO length and the ratio, we can say that the FO length is twice of the ratio of DM.
Therefore, the length of OC will be twice of the ratio of MA.
Length of OC = 10 ft.
Total length = FO + OC = 6+10 = 16 ft.
Volume of the triangular prism = Area of triangular base × length
Volume of the triangular prism = 24×16
Volume of the triangular prism = 384 cu.ft

Question 18.
The volume of the rectangular prism shown below is 2,880 cubic inches. The cross-section shown is parallel to a base. The area of the cross-section is 180 square inches. The length of \(\overline{A B}\) is x inches, and the length of \(\overline{B C}\) is 4x inches.
a) Find the length of \(\overline{A C}\).
Answer:
Given that: The volume of the rectangular prism is 2,880 cubic inches and the area of the cross-section is 180 square inches.
The length of AB is x inches and BC is 4x inches.
Length of AC will be AB+BC = x+4x = 5x

b) Find the value of x.

Answer:
Volume of the rectangular prism = 2,880 cubic inches
Volume of the rectangular prism = Area of rectangular base × length
2,880 = 180 × 5x
5x = 2880 ÷ 180
5x = 16
x = \(\frac{16}{5}\)
Length of AB will be \(\frac{16}{5}\) in.
Length of BC will be 4×\(\frac{16}{5}\) = \(\frac{64}{5}\) in.
Length of AC will be \(\frac{80}{5}\) = 16 in.

Question 19.
In the diagram of a cube shown below, points A, B, C, and D are vertices. Each of the other points on the cube is a midpoint of one of its sides. Describe a cross-section of the cube that will form each of the following figures.
a) a rectangle
b) an isosceles triangle
c) an equilateral triangle
d) a parallelogram

Answer:
a) a rectangle
By joining the CDLJ points, a rectangular cross section can be formed.
b) an isosceles triangle
By joining the MHL points, an isosceles triangle cross section can be formed.
c) an equilateral triangle
By joining the CDM points,an equilateral triangle cross section can be formed.
d) a parallelogram
By joining the MHGL points, a parallelogram cross section can be formed.

Solve. Use graph paper.

Question 20.
Points A, B, C, and D form a square. The area of the square is 9 square units.
a) Find the side length of square ABCD.
Answer:
Given that the area of the square is 9 square units.
Area of square = length × length
9 = length × length
3 × 3 = length × length
Length of the square will be 3 units.

b) The coordinates of point A are (2, 6). Points B and C are below \(\overline{A D}\). Point B is below point A, and point D is to the right of point A. Plot the points in a coordinate plane. Connect the points in order to draw square ABCD.
Answer:
The coordinates of point A are (2, 6).
Point B and C are below point A, so they will be (0,2) and C will be (0,8).
D will be (8,6)

c) The points E, F, G, and H also form a square that is the same size as square ABCD. Point E is 4 units to the right of point A, and 3 units up. Points F and G are below \(\overline{E H}\). Point F is below point E, and point H is to the right of point E. Plot the points in the coordinate plane. Draw \(\overline{E H}\) and \(\overline{G H}\) with solid lines, and \(\overline{E F}\) and \(\overline{F G}\) with dashed lines.
Answer:
Point E is 4 units to the right of point A, and 3 units up. So, it will be (6,9).
Points F and G are below, so they will be (6,3) and (12,3).
Point H will be (12,9).

d) Draw \(\overline{A E}\), \(\overline{D H}\), and \(\overline{C G}\) with solid lines, and \(\overline{B F}\) with a dashed line, Use the solid and dashed lines to see the figure as a solid. Name the type of prism formed.
Answer:
After joining AE, DH, CG and BF.Below prism will be formed.
Since all the sides are of same length, it will form a cube.

e) If the height of the prism is 7 units, find the volume of the prism.
Answer:
Area of the base 7×7 = 49 sq.units
Volume of the prism = 49×7 = 343 cu.units

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.2 Surface Area of Solids detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.2 Answer Key Surface Area of Solids

Math in Focus Grade 6 Chapter 12 Lesson 12.2 Guided Practice Answer Key

Complete.

Question 1.
A cube has edges measuring 6 centimeters each. Find the surface area of the cube.

Area of one square face = •
= cm²
Surface area of cube = •
= cm²

Answer:
Each side of a square measures 6cm.
Area of a square base = length×length
Area of one square face = 6×6 = 36 sq.cm
Surface area of cube = Area of base × height
Surface area of cube = 36×6 = 216 sq.cm

Question 2.
A rectangular prism measures 7 inches by 5 inches by 10 inches. Find the surface area of the prism.

Total area of two orange and two purple faces = ( + + + ) •
= •
= in²
Area of two green rectangular bases = 2 • •
= in²
Surface area of rectangular prism = +
= in²

Answer:
Total area of two orange and two purple faces = (length + width + length + width)× height
Total area of two orange and two purple faces = (5+7+5+7)×10 = 24×10 = 240 sq.in
Area of rectangular base = length×width
Area of two green rectangular bases = 2×(5×7) = 70 sq.in
Surface area of rectangular prism = 240+70= 310 sq.in

Question 3.
The triangular prism shown has three rectangular faces. Its bases are congruent right triangles. Find the surface area of the triangular prism.

Total area of orange, purple, and yellow rectangles = ( + + ) •
= •
= cm²
Area of two green triangular bases = 2 • (\(\frac{1}{2}\) • • ) = cm²
Surface area of triangular prism = + = cm²
Answer:
Total area of orange, purple, and yellow rectangles = (length of three rectangles) × width
Total area of orange, purple, and yellow rectangles = (5+13+12)×9 = 30×9 = 270 sq.cm
Area of a triangle = \(\frac{1}{2}\) × base × height
Area of two green triangular bases = 2×(\(\frac{1}{2}\)×12×13) = 156 sq.cm
Surface area of triangular prism = 270+156 = 426 sq.cm

Question 4.
Alicia makes a pyramid that has an equilateral triangle as its base. The other three faces are congruent isosceles triangles. She measures the lengths shown on the net of her pyramid. Find the surface area of the pyramid.

Area of yellow triangle = \(\frac{1}{2}\) • • = in.²
Area of three blue triangles = 3 • \(\frac{1}{2}\) • • = in.²
Surface area of triangular pyramid = + = in.²
Answer:
Area of a triangle = \(\frac{1}{2}\) × base × height
Area of yellow triangle = \(\frac{1}{2}\) × 5.2 × 6 = 15.6 sq.in
Area of three blue triangles = 3 × \(\frac{1}{2}\) × 10 × 6 = 90 sq.in
Surface area of triangular pyramid = 15.6 + 90 = 105.6 sq.in

Math in Focus Course 1B Practice 12.2 Answer Key

Solve.

Question 1.
A cube has edges measuring 6 centimeters each. Find the surface area of the cube.
Answer:
Measurement of each side of a cube is 6 cm.
Area of a square base= length×length
Surface area of one sqaure side will be 6×6 = 36 sq.cm
Surface area of a cube = 6×Area of base
A cube has 6 sides. Therefore, the total surface area will be 6×36 = 216 sq.cm
The surface area of cube will be 216 sq.cm.

Question 2.
The edge length of a cube is 3.5 inches. Find the surface area of the cube.
Answer:
Measurement of each side of a cube is 3.5 in.
Area of a square base= length×length
Surface area of one sqaure side will be 3.5×3.5 = 12.25 sq.in
Surface area of a cube = 6×Area of base
A cube has 6 sides. Therefore, the total surface area will be 6×12.25 = 75 sq.in
The surface area of cube will be 75 sq.in.

Question 3.
A closed rectangular tank measures 12 meters by 6 meters by 10 meters. Find the surface area of the tank.
Answer:
Total area of four rectangular faces = (length + width + length + width)× height
Total area of four rectangular faces = (12+6+12+6)×10 = 36×10 = 360 sq.m
Area of rectangle = length × width
Area of two rectangular bases = 2×(6×12) = 144 sq.m
Surface area of rectangular prism = 360+144= 504 sq.m

Question 4.
A closed rectangular tank has a length of 8.5 feet, a width of 3.2 feet, and a height of 4.8 feet. Find the surface area of the tank.
Answer:
Total area of four rectangular faces = (length + width + length + width)× height
Total area of four rectangular faces = (8.5+3.2+8.5+3.2)×4.8 = 23.4×4.8 = 112.32 sq.feet
Area of rectangle = length × width
Area of two rectangular bases = 2×(8.5×3.2) = 54.4 sq.feet
Surface area of rectangular prism = 112.32+54.4= 166.72 sq.feet

Question 5.
A triangular prism with its measurements is shown. Find the surface area of the prism.

Answer:
Area of rectangle = length × width
Two rectangle measures 13 cm wide and 15 cm long. So, there area will be 2×13×15 = 390 sq.cm
Area of the third rectangle will be 15×10 = 150 sq.cm
Total area of three rectangular faces = 390+150 = 540 sq.cm
Area of triangle = \(\frac{1}{2}\)×base×height
Area of two triangular bases = 2×\(\frac{1}{2}\)×10×12 = 120 sq.cm
Surface area of triangular prism = 540+120 = 660 sq.cm

Question 6.
A triangular prism with its measurements is shown. Find the surface area of the prism.

Answer:
Area of rectangle = length × width
Area of three rectangular faces = 3×6×4.5 = 81 sq.in
Area of triangle = \(\frac{1}{2}\)×base×height
Area of two triangular bases = 2×\(\frac{1}{2}\)×6×5.2 = 31.2 sq.in
Surface area of triangular prism = 81+31.2 = 112.2 sq.in

Solve.

Question 7.
A square pyramid has four faces that are congruent isosceles triangles. Find the surface area of the square pyramid if the base area is 169 square centimeters.

Answer:
The base area of the given figure is 169 sq.cm.
Since it is a square pyramid, the area will be side×side = 169 sq.cm
169 can also be written as 13 × 13
Therefore,  side × side = 13 × 13
Each side will be 13 cm.
Area of triangle = \(\frac{1}{2}\)×base×height
Area of four isosceles triangles faces = 4×\(\frac{1}{2}\)×13×15 = 390 sq.cm
Total surface area will be 169+390 = 559 sq.cm

Question 8.
The faces of this solid consist of four identical trapezoids and two squares. The side lengths of the two squares are 4 centimeters and 8 centimeters. The height of each trapezoid is 12 centimeters. Find the surface area of the solid.

Answer:
Area of square = length×length
Area of first square = 4×4 = 16 sq.cm
Area of second square will be = 8×8 = 64 sq.cm
Total area of two squares base = 16+64 = 80 sq.cm
The formula to find the area of trapezium = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of four identical trapezoids = 4×\(\frac{1}{2}\) × (4+8) × 12
= 2 × 12 × 12
= 288 sq.cm
The surface area of the solid will be 80+288 = 368 sq.cm

Surface Area of a Trapezoidal Prism = h (b + d) + l (a + b + c + d) square units

Question 9.
Ms. Jones wants to paint the walls of a rectangular room. The height of the room is 8 feet. The floor is 10.5 feet wide and 12 feet long. The doors and windows total 24 square feet and are not going to be painted. Find the total area of the walls that need to be painted.
Answer:
The height of the room is 8 feet and the floor is 10.5 feet wide and 12 feet long.
Surface area of rectangular prism = 2×(length×width + width×height + height×length)
Area of the room will be = 2×(8×10.5 + 8×12 + 12×10.5)
168+192+252 = 612 sq.feet
The doors and windows total 24 square feet and are not going to be painted.As the doors and windows need not be painted, we can remove this area from the total area.
Therefore, the total area of the walls that need to be painted will be 612-24 = 588 sq.feet

Question 10.
The base of a prism has n sides. Write an expression for each of the following.
a) the number of vertices
Answer:
The number of vertices will be 2n.

b) the number of edges
Answer:
The number of edges will be 3n.

c) the number of faces
Answer:
The number of faces will be n+2

Question 11.
The base of a pyramid has m sides. Write an expression for each of the following.
a) the number of vertices
Answer:
The number of vertices will be m+1

b) the number of edges
Answer:
The number of edges will be 2m

c) the number of faces
Answer:
The number of faces will be m+1

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 13 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 13 Review Test Answer Key

Concepts and Skills

Copy and complete the table. Use the set of data.

Question 1.
The number of bedrooms in the units of a new apartment building ranged from 1 to 5. The number of bedrooms in each unit is as follows:

Tabulate the data.

Answer:

Explanation:
A Dataset is a set or collection of data.
This set is normally presented in a tabular pattern.
The number of bedrooms in the units of a new apartment building are marked in tally format.
The number of bedrooms in each unit is value noted in frequency in the above table.

Draw a dot plot and a histogram for the set of data. Include a title.

Question 2.
The table below shows the number of hours 30 teachers in a school spent correcting students assignments.

Answer:
Dot plot for the given above table data,

The histogram shows the ages of runners in a marathon. Use the histogram to answer questions 3 and 4.

Explanation:
Given the table below which shows the number of hours 30 teachers in a school spent correcting students assignments.

A dot plot is a graphical display of data using dots as shown below,

The histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.
The below histogram shows the ages of runners in a marathon.

Question 3.
How many data values are there?
Answer:
25

Explanation:
Given the table below which shows the number of hours 30 teachers in a school spent correcting students assignments.

0 to 9 = 1
10 to 19 = 7
20 to 29 = 11
30 to 39 = 5
40 to 49 = 0
50 to 59 = 1
1 + 7 + 11 + 5 + 0 + 1 = 25

Question 4.
Briefly describe the distribution including any outliers in the data.
Answer:
Yes, 11 is outlier.

Explanation:
Given the table below which shows the number of hours 30 teachers in a school spent correcting students assignments.

1, 1, 5, 11
average of 1,1 is 1
average of 5, 11 is 8
range of 1, 8 is 7
8 + 1.5(7) = 18.5
1 + 1.5(7) = 11.5
As, 11 is less then 11.5
So, 11 is outlier.

Outliers should be investigated carefully. Often they contain valuable information about the process under investigation or the data gathering and recording process.

Problem Solving

The data show the lengths (in inches) of 50 trout caught in a lake during a fishing competition. Use the data to answer questions 5 and 6.

Question 5.
Group the data into suitable intervals and tabulate them. Explain your choice of interval.
Answer:

Explanation:
The above group of data is tabulated with an intervals of 2 inches difference, total 50 fishes are  tabulated in the table as shown in the above figure.
The range is 18 – 8 = 10
Choosing an interval of 2 gives six intervals for the data set.
A larger interval will give too few intervals, and will not be able to see the distribution accurately.

Question 6.
Draw a histogram using the interval. Briefly describe the data.
Answer:

Explanation:
During a fishing competition different trots are caught in the lake.
The lengths (in inches) of 50 trout are tabulated in the table with different length intervals.
The range is 18 – 8 = 10
Choosing an interval of 2 gives six intervals for the data set.
A larger interval will give too few intervals, and will not be able to see the distribution accurately.

The data show the distances a golfer hits (in yards) in a long drive championship. Use the data to answer questions 7 and 8.

Question 7.
Group the data into suitable intervals and tabulate them. Explain your choice of interval.
Answer:

Explanation:
The above data show the distances a golfer hits (in yards) in a long drive championship are tabulated with an interval of 10 yards interval.
The range is 278 -240 = 33
Choosing an interval of 10 gives four intervals for the data set.
A larger interval will give too few intervals, and will not be able to see the distribution accurately.

Question 8.
Draw a histogram using the interval. Briefly describe the data.
Answer:

Explanation:
The above data show the distances of a golfer hits (in yards) in a long drive championship are tabulated with an interval of 10 yards interval.
In horizontal axis line distance and on vertical axis line number of hits are recorded.

The table shows the number of cars passing a traffic light during peak hours on a Friday morning. Use the data to answer questions 9 to 11.

Question 9.
How many cars were observed altogether?
Answer:
Total 240 cars passing a traffic light during peak hours on a Friday morning.

Explanation:
At different time intervals cars passing a traffic light during peak hours on a Friday morning are tabulated in the table. The sum of all cars are as given below.
22 + 45 + 64 + 57 + 27 + 25 = 240
Total 240 cars passing a traffic light during peak hours on a Friday morning

Question 10.
Draw a histogram to display the data.
Answer:

Explanation:
The above histogram shows time on horizontal line and number of cars on the vertical line.
At different time intervals cars passing a traffic light during peak hours on a Friday morning are tabulated in the table.
Total 240 cars passing a traffic light during peak hours on a Friday morning are shown in the above histogram diagram.

Question 11.
Describe the distribution of the data. Suggest why the histogram has the shape that it does.
Answer:
The distribution of a data set is the shape of the graph when all possible values are plotted on a frequency graph showing how often they occur.
Usually, we are not able to collect all the data.
Therefore we take a random sample.
This sample is used to make conclusions about the whole data set.

Explanation:
The above histogram shows time on horizontal line and number of cars on the vertical line.
At different time intervals cars passing a traffic light during peak hours on a Friday morning are tabulated in the table.
Total 240 cars passing a traffic light during peak hours on a Friday morning are shown in the above histogram diagram.

The quiz scores of 94 students are shown in the table. Use the data to answer questions 12 to 14.

Question 12.
Find the value of x.
Answer:
x = 7

Explanation:
The given table shows the quiz scores of 94 students.
To find x, first find the sum of the number of students who scored in quiz.
17 + 3 + 5 + 12 + 15 + 17 + 10 + 8 + x = 94
87 + x = 94
x = 94 – 87
x = 7

Question 13.
Draw a histogram to represent the data. Describe the distribution of the data.
Answer:

Explanation:
The distribution of a data set is the shape of the graph when all possible values are plotted on a frequency graph showing how often they occur in the above.
Usually, we are not able to collect all the data.
Therefore we take a random sample.
This sample is used to make conclusions about the whole data set.

Question 14.
If the 5 students who scored 23-25 all scored 26 instead, would this change where most of the data occur? Justify your answer.
Answer:
The changes of data table and the histogram is plotted for the changed values are shown in the below explanation.

Explanation:

If the 5 students who scored 23-25 all scored 26 instead,
the value 5 in the 23-25 is shifted to 26-28 scores student, as shown below.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 13 Lesson 13.3 Histograms detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 13 Lesson 13.3 Answer Key Histograms

Math in Focus Grade 6 Chapter 13 Lesson 13.3 Guided Practice Answer Key

Draw a histogram for each set of data.

Question 1.
In a study of the length of several individuals of one species of fish caught, the following observations were recorded. The lengths were measured to the nearest centimeter.

Answer:

Explanation:
A histogram is an approximate representation of the distribution of numerical data.
As shown in the above table of data the fishes are of different lengths are graphically represented.
In statistics, a histogram is a graphical representation of the distribution of data.
The histogram is represented by a set of rectangles, adjacent to each other,
where each bar represent a kind of data.

Question 2.
The scores obtained by 40 students in a mathematics quiz are recorded in the table below.

Answer:

Explanation:
We know that in statistics, a histogram is a graphical representation of the distribution of data.
The histogram is represented by a set of rectangles, adjacent to each other,
where each bar represent a kind of data.
A histogram is an approximate representation of the distribution of numerical data.
As shown in the above table of data the scores obtained by 40 students in a mathematics quiz.

Draw a histogram for each set of data. Solve.

Question 3.
The cholesterol levels (in milligram per deciliter) of 40 men are listed below.

a) Group the data into suitable intervals and tabulate them. Explain your choice of intervals.
Answer:

Explanation:
Given that, The cholesterol levels (in milligram per deciliter) of 40 men are listed below.

The cholesterol levels (in milligram per deciliter) of 40 men are listed above table with an interval of 50 points.
As shown in the below table 100 – 150, 151 – 200 …… 351 – 400.

b) Draw a histogram using the interval.
Answer:

Explanation:
The histogram diagram for the cholesterol levels (in milligram per deciliter) of 40 men are listed above table with an interval of 50 points as 100 – 150, 151 – 200 …… 351 – 400 and,
minimum of 1 man and maximum of 23 men histogram are plotted in the above diagram.

Question 4.
The speeds in kilometers per hour of 40 cars on a highway were recorded as follows.

a) Group the data into suitable intervals and tabulate them. Explain your choice of intervals.
Answer:

Explanation:
As we know that grouping data means the data given in the form of class intervals such as 0-10 to 91-100 as shown above.
Given that the speeds in kilometers per hour of 40 cars on a highway were recorded with an intervals of 10 kmph difference as shown in the above table.

b) Draw a histogram using the interval.

Answer:
Total 40 cars of different speed histogram diagram

Explanation:
A histogram is a graphical representation of discrete or continuous data.
The area of a bar in a histogram is equal to the frequency.
The y -axis is plotted by frequency density and the x -axis is plotted with the range of values divided into intervals.
Given different care on highway with different speed is recorded in the table with different speed intervals as shown in the above histogram is plotted.

Describe the data.

Question 5.
The histogram shows the number of representatives each state sent to the U.S. Congress in 2011. Briefly describe the data.

Answer:

Explanation:
The histogram shows the number of representatives each state sent to the U.S. Congress in 2011 describe the data in table form.
A histogram is a graphical representation of discrete or continuous data.
The y -axis is plotted by frequency density and the x -axis is plotted with the range of values divided into intervals.
The above table consists of Number of representatives and Number of states and the values with reference to the histogram is tabulated.

Question 6.
The histogram shows the highest temperature (in degrees Fahrenheit) recorded during December for one city. The temperature were recorded to the nearest degree. Briefly describe the data.

Answer:

Explanation:
The histogram shows the highest temperature (in degrees Fahrenheit) recorded during December for one city.
A histogram is a graphical representation of discrete or continuous data.
The area of a bar in a histogram is equal to the frequency.
The y -axis is plotted by frequency density and the x -axis is plotted with the range of values divided into intervals.
So, the temperature were recorded to the nearest degree.
The data is tabulated in the table format.

Math in Focus Course 1B Practice 13.3 Answer Key

Draw a histogram for each set of data. Include a title.

Question 1.
The table shows the number of cans recycled by 25 households in a month.

Answer:

Explanation:
The histogram shows the number of cans recycled by 25 households in a month, data given in the table format.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

Question 2.
The table shows the number of points scored by a football team in 20 games of one season.

Answer:

Explanation:
The histogram shows the number of points scored by a football team in 20 games of one season.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

Question 3.
The table shows the keyboarding speed of 100 students in a beginning keyboarding class.

Answer:

Explanation:
The histogram shows the keyboarding speed of 100 students in a beginning keyboarding class.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

Draw a histogram for the set of data. Include a title. Solve.

Question 4.
The number of sunny days in a year for 200 cities are shown in the table.

a) Find the value of x.
Answer:
48

Explanation:
The sum of the cities in the table are,
21 + 25 + 32 + 31 + 24 + 19 = 152
There must be 200 cities,
then 200 – 152 = 48
x = 48

b) Draw a histogram to represent the data. Briefly describe the data.
Answer:

Explanation:
The histogram shows, the number of sunny days in a year for 200 cities.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

c) What percent of the cities had fewer than 149 sunny days?
Answer:
10.5%

Explanation:
Total number of cities 200.
The number of cities had fewer than 149 sunny days = 21
The percent of the cities had fewer than 149 sunny days,
x = (21 / 200) X 100
x = 10.5%

d) What percent of the cities had more than 184 sunny days?
Answer:
21.5%

Explanation:
Total number of cities 200.
the number of cities had more than 184 sunny days,
24 + 19 = 43
The percent of the cities had more than 184 sunny days,
x = (43 / 200) X 100
x = 21.5%

The histogram shows the number of cars observed at one intersection at different times of the day. Use the histogram to answer questions 5 to 9.

Question 5.
How many observations are there?
Answer:
820 observations at different time intervals.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.
Number of observations are as follows,
180 + 110 + 130 + 100 + 80 + 220 = 820

Question 6.
How many fewer cars passed the intersection from 6 A.M. to 7:59 A.M. than from 4 P.M. to 5:59 P.M.?
Answer:
40 observations.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
cars passed the intersection from 6 A.M. to 7:59 A.M =180
cars passed the intersection from 4 P.M. to 5:59 P.M = 220
220 – 180 = 40
40 observations shows fewer cars passed the intersection from 6 A.M. to 7:59 A.M. than 4 P.M. to 5:59 P.M.

Question 7.
How many more cars passed the intersection from 10 A.M. to 11:59 A.M. than from 2 P.M. to 3:59 P.M.?
Answer:
50 cars.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
cars passed the intersection from 10 A.M. to 11:59 A.M =130
cars passed the intersection from 2 P.M. to 3:59 P.M = 80
So, 130 – 80 = 50
50 cars passed the intersection from 10 A.M. to 11:59 A.M. than from 2 P.M. to 3:59 P.M.

Question 8.
What percent of the number of cars that passed the intersection from 4 P.M. to 5:59 P.M. was observed from 8 A.M. to 9:59 A.M.?
Answer:
50 percent of the number of cars that passed the intersection from 4 P.M. to 5:59 P.M. was observed from 8 A.M. to 9:59 A.M.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
cars passed the intersection from 8 A.M. to 9:59 A.M =110
cars passed the intersection from 4 P.M. to 5:59 P.M = 220
220 – 110 = 110
110 / 220 = 0.5
0.5 x 100 = 50%
50 percent of the number of cars that passed the intersection from 4 P.M. to 5:59 P.M. was observed from 8 A.M. to 9:59 A.M.

Question 9.
What percent of the total number of cars that passed the intersection from 6 A.M. to 5:59 P.M. was observed from 4 P.M. to 5:59 P.M.? Round your answer to the nearest percent.
Answer:
73%

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
The total number of cars that passed the intersection from 6 A.M. to 5:59 P.M = 820
The total number of cars that passed the intersection from 4 P.M. to 5:59 P.M = 220
820 – 220 = 600
(600/820)X 100 = 73.17%
Round the answer to the nearest percent 73%
73% percent of the total number of cars that passed the intersection from 6 A.M. to 5:59 P.M. was observed from 4 P.M. to 5:59 P.M.

The histogram shows the number of books students in a class read last month. Use the histogram to answer the question.

Question 10.
Briefly describe the data. Explain whether the histogram shows any outlier of the data set.
Answer:
Yes, 16- 20 books is outlier of the data set.

Explanation:
The above histogram shows the number of books students in a class read last month.
The gap of  no books reading between 16 – 20.
An outlier is an observation that lies an abnormal distance from other values in a random sample from a population.
In a sense, this definition leaves it up to the analyst to decide what will be considered abnormal. Before abnormal observations can be singled out, it is necessary to characterize normal observations.

Draw a histogram for the set of data. Include a title. Solve.

Question 11.
The sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.

2 pairs of shoes of at least size 9.5 were sold.
a) Find the values of x and y.
Answer:
x = 1
y = 2

Explanation:
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the above table.
The sales figures for 60 pairs of one style of shoe of various sizes are,
8 + 22 + 16 + 4 + 3 + 3 + 1 + x + y = 60
57 + x + y = 60
x + y = 60 – 57
x + y = 3
2 pairs of shoes of at least size 9.5 were sold.
y = 2
x + 2 = 3
x = 3 – 2 = 1
x = 1

b) What fraction of the shoes sold are smaller than size 8?
Answer:
0.783

Explanation:
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.
The shoes sold are smaller than size 8 = 1 + 8 + 22 + 16 = 47
Totals shoes sold are = 60
The fraction of the shoes sold are smaller than size 8.
47/60 = 0.783

c) Draw a histogram using the intervals 6-6.5, 7-7.5, 8-8.5, and so on. Briefly describe the data.
Answer:

Explanation:
The histogram diagram shown above is the number of shoes sold of different sizes.
on vertical number of shoes sold and on horizontal line the sizes of the shoes sold.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

d) If shoes were to be categorized as follows:
small – sizes 6 to 7; medium – sizes 7.5 to 8.5; large – sizes 9 to 10, draw a histogram for the above data using the new categories.
Answer:

Explanation:
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.
So, shoes were to be categorized as small – sizes 6 to 7; medium – sizes 7.5 to 8.5; large – sizes 9 to 10, and the quantities are added to total for small, medium and large.
Number of small shoes are 31
Number of medium shoes are 23
Number of large shoes are 6 pairs.
The total number of shoes are = 31 + 23 + 6 = 60

e) Compare the two histograms. When would each one be more useful?
Answer:
Both are useful,
for exact sale with reference to size first histogram are very useful.
second histogram tells about the small or medium or large size shoes are sold more or less in quantity.

Explanation:
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.
for exact sale with reference to size first histogram are very useful.
second histogram tells about the small or medium or large size shoes are sold more or less in quantity as shoes were to be categorized as small – sizes 6 to 7; medium – sizes 7.5 to 8.5; large – sizes 9 to 10, and the quantities are added to total for small, medium and large.

Question 12.
Math Journal A survey was carried out to find the number of players who scored a certain number of goals during soccer matches in a month. A histogram was drawn to display the results.

Is the histogram drawn correctly? Discuss with your partner and explain your thinking.
Answer:
The histogram is not drawn correctly.

Explanation:
As per the histogram diagram, a survey was carried out to find the number of players who scored a certain number of goals during soccer matches in a month.
No values in histogram are recorded on horizontal line.

Brain @ Work

The table below shows the test scores for all the students in a Spanish I course.

The following grades are used to represent the scores.

Question 1.
Draw a histogram to show the distribution of the scores.
Answer:

Explanation:
Based on the table below shows the test scores for all the students in a Spanish I course.

The above histogram on vertical line frequency is recorded and,
on the horizontal line scores are marked.
Bar graphs are plotted corresponding values given in the table.

Question 2.
Draw a bar graph to display the grades.
Answer:

Explanation:
The following grades are used to represent the scores on histogram as shown above.

In the above histogram on vertical line Grades are marked and on the horizontal line scores are recorded.
Bar graphs are plotted corresponding grades given in the table.

Question 3.
Five students increased their scores from 79 to 89.
a) How would it change the histogram?
Answer:

Explanation:
In the above histogram on vertical line new frequency is recorded after shifting the five students from 79 to 89 scored students and on the horizontal line scores are marked.
Bar graphs are plotted corresponding values given in the table.

five students increased their scores from 79 to 89,
it means that subtract fives students from 73 – 79 and,
add it to 80 – 89 scores as shown in the above table.

b) How would it change the bar graph?
Answer:
Corresponding graph also will change accordingly as shown.

Explanation:
In the above histogram on vertical line new frequency is recorded after shifting the five students from 79 to 89 scored students and on the horizontal line scores are marked.
Bar graphs are plotted corresponding values given in the table.
Five students increased their scores from 79 to 89,
it means the subtract fives students from 73 – 79 and,
add it to 80 – 89 scores as shown in the above table.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 13 Lesson 13.1 Collecting and Tabulating Data detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 13 Lesson 13.1 Answer Key Collecting and Tabulating Data

Math in Focus Grade 6 Chapter 13 Lesson 13.1 Guided Practice Answer Key

Complete. Use the data in the table.

Emma used a questionnaire to find out the number of brothers or sisters her classmates have in their families.

Then she used a tally chart to record what she had found.

Question 1.
Copy and complete the table by counting the tally marks.

Answer:

Explanation:
Give that,
Emma used a questionnaire to find out the number of brothers or sisters her classmates have in their families.
Then she used a tally chart to record what she had found.
We know that the tally marks are a way to mark or record your counting.
Tally marks are a numerical system used to make counting easier.
As the name suggests, it is a system that helps keep the “tally” of things by number.
Tally marks are commonly used for counting scores, points, number of people, etc.
Tally marks differ from country to country, as each culture has developed its own systems.

Question 2.
More of Emma’s classmates have brother or sister than any other number.
Answer:
1

Explanation:
Given that, Emma used a questionnaire to find out the number of brothers or sisters her classmates have in their families.
Then she used a tally chart to record what she had found.
with reference to the above Tally marks and frequency table,
Emma’s classmates have 1 brother or sister than any other number.

Question 3.
of Emma’s classmates have 3 brothers or sisters in their families.
Answer:
5 of Emma’s classmates have 3 brothers or sisters in their families.

Explanation:
Given that, Emma used a questionnaire to find out the number of brothers or sisters her classmates have in their families.
Then she used a tally chart to record what she had found.
with reference to the above Tally marks and frequency table,
5 of Emma’s classmates have 3 brothers or sisters in their families.

Question 4.
of Emma’s classmates have 4 brothers or sisters in their families.
Answer:
2 of Emma’s classmates have 4 brothers or sisters in their families.

Explanation:
Given that, Emma used a questionnaire to find out the number of brothers or sisters her classmates have in their families.
Then she used a tally chart to record what she had found.

with reference to the above Tally marks and frequency table,
2 of Emma’s classmates have 4 brothers or sisters in their families.

Question 5.
How many more of Emma’s classmates have 2 brothers or sisters than 5 or more brothers or sisters in their families?
Answer:
6 of Emma’s classmates have 2 brothers or sisters than 5 or more brothers or sisters in their families.

Explanation:
Given that, Emma used a questionnaire to find out the number of brothers or sisters her classmates have in their families.
Then she used a tally chart to record what she had found.
7 of Emma’s classmates have 2 brothers or sisters in their families
1 of Emma’s classmates have more then 5 brothers or sisters in their families
if we subtract as follows
7 – 1 = 6
6 many more of Emma’s classmates have 2 brothers or sisters than 5 or more brothers or sisters in their families.

Question 6.
Emma has classmates altogether.
Answer:
27

Explanation:
Given that, Emma used a questionnaire to find out the number of brothers or sisters her classmates have in their families.
Then she used a tally chart to record what she had found.

So, Emma has 27 classmates altogether.

Hands-On Activity

COLLECT, TABULATE, AND INTERPRET DATA

Materials:

• blank table
• ruler
• collection of pencils

Work in groups of three or four.

Step 1: Collect a set of pencils of various lengths. Use a ruler to measure the length of each pencil to the nearest centimeter. Use tally marks to record the data.
Step 2: Tally your results on a table like the one below. Then count the tally marks to complete the table.

Step 3: Write at least four questions about the data in the table using these phrases.

Answer:

Explanation:
We know that the tally marks are a way to mark or record your counting.
Tally marks are a numerical system used to make counting easier.
As the name suggests, it is a system that helps keep the “tally” of things by number.
Tally marks are commonly used for counting scores, points, number of people, etc.
Tally marks differ from country to country, as each culture has developed its own systems.

Write at least four questions about the data in the above table.

1) How many pencils are of shortest lengths?
Answer:
5 pencils.

Explanation:
Given that to collect a set of pencils of various lengths.
Use a ruler to measure the length of each pencil to the nearest centimeter.
Use tally marks to record the data.
Tally your results on a table like the one above.
Then count the tally marks to complete the table.
with reference to the above table, there are 5 pencils of length 14 cm are shortest.

2) How many pencils are of longest lengths?
Answer:
2 pencils.

Explanation:
Given that to collect a set of pencils of various lengths.
Use a ruler to measure the length of each pencil to the nearest centimeter.
Use tally marks to record the data.
Tally your results on a table like the one above.
Then count the tally marks to complete the table.
with reference to the above table, there are 2 pencils are of length 18 cm are longest.

3) How many pencils are of more number of which lengths?
Answer:
10 pencils.
Explanation:
Given that to collect a set of pencils of various lengths.
Use a ruler to measure the length of each pencil to the nearest centimeter.
Use tally marks to record the data.
Tally your results on a table like the one above.
Then count the tally marks to complete the table.
with reference to the above table, there are 10 pencils are of length 16 cm long.

4) Altogether total how many pencils are of different lengths?
Answer:
28 pencils.

Explanation:
Given that to collect a set of pencils of various lengths.
Use a ruler to measure the length of each pencil to the nearest centimeter.
Use tally marks to record the data.
Tally your results on a table like the one above.
Then count the tally marks to complete the table.
with reference to the above table, there are 28 pencils are of different lengths.

Math in Focus Course 1B Practice 13.1 Answer Key

Copy and complete the table. Solve.

Question 1.
A shampoo company wanted to find out more about its customers. So they asked 30 customers to indicate their income bracket:
Below $500 per week
$500—$1,000 per week
Over $1,000 per week
A tally chart was used to record the findings.

How many customers have a weekly income of $1,000 or less?
Answer:
26 customers.

Explanation:
Given that, A shampoo company wanted to find out more about its customers.
So they asked 30 customers to indicate their income bracket:
Below $500 per week
$500—$1,000 per week
Over $1,000 per week
A tally chart was used to record the findings.
with reference to the table below,
the total number of customers have a weekly income of $1,000 or less are 26.
customers have a weekly income of $500 – $1,000 is 19 and
customers have a weekly income of $500 are 7
19 + 7 = 26 customers

Question 2.
Fifty students were asked their level of satisfaction with the school’s music program. The following responses were the choices provided:
(a) very satisfied (b) satisfied (c) neutral (d) dissatisfied (e) very dissatisfied.

a) How many students are satisfied or very satisfied?
Answer:
students are satisfied = 7
students are very satisfied = 2

Explanation:
Given that, Fifty students were asked their level of satisfaction with the school’s music program.
The following responses were the choices provided:
(a) very satisfied (b) satisfied (c) neutral (d) dissatisfied (e) very dissatisfied.

With reference to the above table, students are satisfied or very satisfied is shown below.
students are satisfied = 7
students are very satisfied = 2

b) Based on the results of the survey, should the school think about changing the program? Explain your reasoning.
Answer:
YES

Explanation:
Given that, Fifty students were asked their level of satisfaction with the school’s music program.
The following responses were the choices provided:
(a) very satisfied (b) satisfied (c) neutral (d) dissatisfied (e) very dissatisfied.
Based on the above table results of the survey,
the school think about changing the program due to less satisfied and very satisfied out of 50 students.

Question 3.
A mathematics teacher wanted to find out how many hours per week his students spent on math homework. The average number of hours reported by each student is shown.
5, 3, 6, 8, 2, 4, 2, 1, 9, 1, 9, 6, 4, 6, 5, 1, 10, 1, 5, 6, 7, 8, 6, 10, 7, 5, 2, 8
Arrange the numbers in ascending order.
1,1, 1, 1, 2, 2, 2, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10

How many students spent more than 3 hours per week on their math homework?
Answer:
20 students spent more than 3 hours and less then per week on their math homework.

Explanation:
Given that, A mathematics teacher wanted to find out how many hours per week his students spent on math homework.
The average number of hours reported by each student is shown.
5, 3, 6, 8, 2, 4, 2, 1, 9, 1, 9, 6, 4, 6, 5, 1, 10, 1, 5, 6, 7, 8, 6, 10, 7, 5, 2, 8
Arrange the numbers in ascending order.
1,1, 1, 1, 2, 2, 2, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10
With reference to the given information,
13 students spent more than 4 hours per week on their math homework.
13 + 7 = 20

Question 4.
Shelly conducted a survey among her friends. She asked them to choose their favorite fruit from this list of fruits: apple, orange, strawberry, grapes, and peach. These are the data she collected:

Tabulate the data.

What is the favorite fruit of Shelly’s friends?
Answer:
Strawberry.

Explanation:
Given that, Shelly conducted a survey among her friends.
She asked them to choose their favorite fruit from this list of fruits: apple, orange, strawberry, grapes, and peach.
These are the data she collected:
With reference to the survey among her friends.
Shelly friend’s favorite fruit from the list of fruits: apple, orange, strawberry, grapes, and peach:
Strawberry is favorite fruit as shown below.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 13 Introduction to Statistics detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 13 Answer Key Introduction to Statistics

Math in Focus Grade 6 Chapter 13 Quick Check Answer Key

Complete. Use the data in the line plot.

The line plot shows the weight, in pounds, of cartons of apples in a store. Each x represents 1 carton of apples.

Question 1.
What is the weight of the lightest carton of apples?
Answer:
45 lb
Explanation:
The line plot shows the weight, in pounds, of cartons of apples in a store.
Each x represents 1 carton of apples.

So, the weight of one apple carton box (x) = 45 lb
= 45 lb

Question 2.
What is the weight of the heaviest carton of apples?
Answer:
378 lb
Explanation:
The line plot shows the weight, in pounds, of cartons of apples in a store.
Each x represents 1 carton of apples.

9 apple carton box X 45 lb.
The weight of the heaviest carton of apples is = 378 lb

Question 3.
What is the difference in weight between the heaviest carton of apples and the lightest carton of apples?
Answer:
333 lb
Explanation:
We know from the above questions that the weight of the heaviest carton of apples = 378 lb
The weight of the lightest carton of apples = 45 lb
The difference in weight between the heaviest carton of apples and the lightest carton of apples,
378 – 45 = 333 lb

Question 4.
How many cartons weigh more than 41 pounds?
Answer:
15 cartons weigh more than 41 pounds.
Explanation:
The line plot shows the weight, in pounds, of cartons of apples in a store.

9 cartons weigh 42 pounds
3 cartons weigh 43 pounds
2 cartons weigh  44 pounds
1 cartons weigh  45 pounds
9 + 3 + 2 + 1 =  15
15 cartons weigh more than 41 pounds.

Question 5.
How many cartons weigh less than 40 pounds?
Answer:
7 cartons weighs less than 40 pounds.
Explanation:
The line plot shows the weight, in pounds, of cartons of apples in a store.

5 cartons weigh  38 pounds.
2 cartons weigh  39 pounds.
5 + 2 =  7
So, 7 cartons weigh less than 40 pounds.

Question 6.
How many cartons weigh 44 pounds each?
Answer:
2 cartons weighs 44 pounds each.
Explanation:
The line plot shows the weight, in pounds, of cartons of apples in a store.
So, 2 cartons weighs 44 pounds each.

Question 7.
How many cartons are there in all?
Answer:
32 cartons are there in all.
Explanation:
The line plot shows the weight, in pounds, of cartons of apples in a store.

So, 32 cartons are there in all.

Question 8.
How many times as many cartons of apples weigh 40 pounds as the number of cartons of apples that weigh 43 pounds?
Answer:
2 times.
Explanation:
cartons of apples weigh 40 pounds is 6 cartons.
cartons of apples that weigh 43 pounds is 3 cartons.
Number of times as many cartons of apples weigh 40 pounds as the number of cartons of apples that weigh 43 pounds = 6/3 = 2 times.

Question 9.
The ratio of the number of cartons of apples that weigh 42 pounds to the ‘ number of cartons of apples that weigh 40 pounds is .
Answer:
3 : 2
Explanation:
cartons of apples weigh 42 pounds is 9 cartons.
cartons of apples that weigh 40 pounds is 6 cartons.
Ratio of both the cartons = 9 : 6
By simplifying we get 3 : 2
So, the ratio of the number of cartons of apples that weigh 42 pounds to the ‘ number of cartons of apples that weigh 40 pounds is 3 : 2

Question 10.
The number of cartons of apples that weigh 41 pounds is % of the total number of cartons of apples.
Answer:
12.5
Explanation:
The number of cartons of apples that weigh 41 pounds is 4 cartons.
The total number of cartons of apples = 32
4/32 = 0.125
So, the number of cartons of apples that weigh 41 pounds is 12.5 % of the total number of cartons of apples.

This handy Math in Focus Grade 6 Workbook Answer Key Cumulative Review Chapters 12-14 detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Cumulative Review Chapters 12-14 Answer Key

Concepts and Skills

Match each of the solid figures to its net. (Lesson 12.1)

Question 1.

Answer:
1 – b
2 – a
3 – c

Explanation:
A triangular prism is a polyhedron made up of two triangular bases and three rectangular sides.
It is a three-dimensional shape that has three side faces and two base faces, connected to each other through the edges.

In geometry, a cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each vertex.

A pyramid is a three-dimensional figure. It has a flat polygon base.
All the other faces are triangles and are called lateral faces.
The number of lateral faces equals the number of sides on its base.
Its edges are the line segments formed by two intersecting faces.

Find the surface area and volume of each prism. (Lesson 12.1, 12.2)

Question 4.

Answer:
Surface Area SA = 1308 square meter.
Volume V = 2808 cubic meters.

Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.
SA=2lw+2lh+2hw
SA = 2x 26×9 + 2x 26×12 + 2x 9x 12
SA = 468 + 624 + 216
SA = 1308 sq meter
Volume of a cuboid:
The volume of cuboid is the quantity that is used to measure the space in a cuboid.
A cuboid is a three-dimensional shape that can be seen around us very often.
The term volume is used in measuring the capacity of any shape based on its dimensions such as: length, breadth, and, height. To calculate the volume of a cuboid,
V = l × b × h = lbh
V = 26 x 9 x 12
V = 2808 cubic meters.

Question 5.

Answer:
Surface Area = 828 square inches.
Volume = 1080 cube inches.

Explanation:
The surface area of triangular prism is the total area of all its faces.
A triangular prism is a prism that has two congruent triangular faces and three rectangular faces that join the triangular faces.
has 6 vertices, 9 edges, and 5 faces. Let us learn more about the surface area of a triangular prism.

Surface area = (Perimeter of the base × Length of the prism) + (2 × Base Area)
= (S₁ +S₂ + S₃)L + bh
=(12 + 15 + 9)20 + 15 x 20
= 720 + 108
= 828 square inches.
Volume:
V =(1/2)( l × b × h) = (0.5)lbh
V = 0.5 x 20 x 9 x 12
V = 1,080 cubic meters.

Solve. Show your work. (Lessons 14.1, 14.2)

Question 6.
The data set shows the lengths (in inches) of seven pieces of wire.
7.9, 6.8, 7.6, 9.9, 10.1,9.1, 10.9
Find the mean and median lengths of these seven pieces of wire.
Answer:
mean = 8.9
median = 9.1

Explanation:
Given set of data, 7.9, 6.8, 7.6, 9.9, 10.1,9.1, 10.9
arrange the given data in the ascending order,
6.8, 7.6, 7.9, 9.1, 9.9, 10.1, 10.9
Mean:
The mean = \(\frac{Sum of a set of items}{Number of items}\)
mean = \(\frac{6.8 + 7.6 + 7.9 + 9.1 + 9.9 + 10.1 + 10.9}{7}\)
mean = \(\frac{62.9}{7}\)
mean = 8.9
Median:
median : Middle value is the median of a given data set.
6.8, 7.6, 7.9, 9.1, 9.9, 10.1, 10.9
the above data has 7 observations,
So, 9.1 in the middle of the order sequence is the median.
median = 9.1

Question 7.
The data set shows the weights (in pounds) of 9 vases.
8.8, 8.3, 7.7, 11.6, 9.9, 8.9, 10.4, 9.6, 8.5
Find the mean and median weights of these 9 vases.
Answer:
Mean = 9.3lb
Median = 8.9lb

Explanation:
Given set of data, 8.8, 8.3, 7.7, 11.6, 9.9, 8.9, 10.4, 9.6, 8.5
arrange the given data in the ascending order,
7.7, 8.3, 8.5, 8.8, 8.9, 9.6, 9.9, 10.4, 11.6
Mean:
The mean = \(\frac{Sum of a set of items}{Number of items}\)
mean = \(\frac{7.7 + 8.3 + 8.5 + 8.8 + 8.9 + 9.6 + 9.9 + 10.4 + 11.6}{9}\)
mean = \(\frac{83.7}{9}\)
mean = 9.3

Median:
median : Middle value is the median of a given data set.
7.7, 8.3, 8.5, 8.8, 8.9, 9.6, 9.9, 10.4, 11.6
the above data has 9 observations,
8.9 in the middle of the order sequence is the median.

Question 8.
The data set shows the heights (in feet) of 8 trees.
53, 56, 65, 61, 67, 60, 52, 48
Find the mean and median heights of these 8 trees.
Answer:
Mean = 57.75
Median = 58

Explanation:
Given set of data, 53, 56, 65, 61, 67, 60, 52, 48
arrange the given data in the ascending order,
48, 52, 53, 56, 60, 61, 65, 67
Mean:
The mean = \(\frac{Sum of a set of items}{Number of items}\)
mean = \(\frac{48 + 52 + 53 + 56 + 60 + 61 + 65 + 67}{8}\)
mean = \(\frac{462}{8}\)
mean = 57.75

Median:
median : Middle value is the median of a given data set.
48, 52, 53, 56, 60, 61, 65, 67
the above data has 8 observations,
So, 56, 60 are in the middle of the order sequence is the median.
the average of (56 + 60)/2 = 116/2 = 58

The volume of each triangular prism is 497 cubic feet. Find the height of the triangular base. Round your answers to the nearest tenth of a foot. (Lesson 12.3)

Question 9.

Answer:
6.9 feet.

Explanation:
Volume:
V =(1/2)( l × b × h) = (0.5) lbh
V = 0.5 x 18 x 8 x h
497 = 72h
h = \(\frac{497}{72}\)
h = 6.9 ft

Question 10.

Answer:
9.94 ft

Explanation:
Volume:
V =(1/2)( l × b × h) = (0.5) lbh
V = 0.5 x 20 x 5 x h
497 = 50h
h = \(\frac{497}{50}\)
h = 9.94 ft

Solve. (Lesson 12.3)

Question 11.
The solid below is made of identical cubes. The volume of the solid is 405 cubic centimeters. Find the edge length of each cube.

Answer:
3 cm

Explanation:
The volume of a cube is length x width x height.
Since it’s a cube, though, the length, width, and height are all equal, and equivalent to the length of one edge of the cube.
Therefore, to find the length of an edge of the cube,
just find the cube root of the volume.
Formula for volume of cube is given by V=a³
where a is edge of the cube.
Given,
volume = 405 cm³
The volume of the solid is 405 cubic centimeters, there are 15 cubes
405 /15 = 27
volume = 27 cm³
a³  = 27
a = 3 cm

Draw a dot plot and a histogram for the set of data. Include a title. (Lessons 13.2, 13.3)

Question 12.
The number of pieces of fruits eaten in the past two days by each of 30 students was recorded below.

a) Represent the set of data with a dot plot.
Answer:

Explanation:
Each dot represent one children,
the number of pieces of fruits eaten in the past two days by each of 30 students was recorded with orange dots.

b) Group the data into suitable intervals and tabulate them.
Answer:

Explanation:
With an interval of one numbers all dots are tabulated in a table as shown in the above table for the above data given.

c) Draw a histogram using the intervals from part b). Briefly describe the data.
Answer:

Explanation:
Histogram drawn using the intervals from the data in the table as shown in the above table.
x-axis shows the number of students and y-axis shows number of pieces of fruits.
With an interval of one number as shown above.
The shape of the histogram is right skewed.

Describe the data. (Lesson 13.3)

Question 13.
The histogram shows the number of floors each building has in a particular city. Briefly describe the data.

Answer:
Total buildings = 84
Most of the buildings have 21 to 40 floors.
The range of the data is 119.
Most of the data values are to the right of the interval 21-40,
The shape of the histogram is right skewed.

Explanation:

The above histogram shows the number of floors each building has in a particular city.
As, total buildings, range of the data, intervals of the data and shape of the histogram are described.

Problem Solving

Draw a dot plot for each set of data. Use your dot plot to answer each question. (Chapters 13, 14)

Question 14.
The data set shows the number of text messages sent by Emily in 14 days.

a) Represent the set of data with a dot plot.
Answer:

Explanation:
The above dot plot shows the data set and the number of text messages sent by Emily in 14 days.. Each dot represents one message.

b) Find the mean, median, and mode of the data set.
Answer:
Mean = 3
Mode = 7
Median = 4

Explanation:
The above dot plot shows the data set and the number of text messages sent by Emily in 14 days.. Each dot represents one message.
Mean:
The mean = \(\frac{Sum of a set of items}{Number of items}\)
mean = \(\frac{0x2 + 1×2 + 2×1 + 3×2 + 5×2 + 7×4 + 8×1 }{14}\)
mean = \(\frac{0 + 2 + 2 + 6 + 10 + 14 + 8 }{14}\)
mean = \(\frac{42}{14}\)
mean = 3

Mode:
Mode is the value which occurs the maximum number of times in a given data set.
It is the third measure of the central tendency.
0, 0, 1, 1, 2, 3, 3, 5, 5, 7, 7, 7, 7, 8
the above data has 14 observations,
So,  7 appears most frequently, is the mode of a given data.

Median:
median : Middle value is the median of a given data set.
0, 0, 1, 1, 2, 3, 3, 5, 5, 7, 7, 7, 7, 8
the above data has 14 observations,
Median is the average value of the 3 & 5
(3 + 5 )/2 = 8/2 = 4
So, 4 in the middle of the order sequence is the median.

Question 15.
The data set shows the number of salads served in a cafe for each of 20 days.

a) Represent the set of data with a dot plot.
Answer:

Explanation:
The given data set shows the number of salads served in a Cafe for each of 20 days.
Each dot represents the number of salad served in Cafe.

b) Find the mean, median, and mode of the data set.
Answer:
Mean = 21.8
Median = 22
Mode = 24

Explanation:
The given data set shows the number of salads served in a Cafe for each of 20 days.
Mean:
The mean = \(\frac{Sum of a set of items}{Number of items}\)
mean = \(\frac{1×16 + 1×17 + 1×18 + 2×19 + 2×20 + 2×21 + 3×22 + 2×23 + 4x 24+ 2×26 }{14}\)
mean = \(\frac{16 + 17 + 18+ 38 + 40 + 42 + 66 + 46 + 96 + 52 }{14}\)
mean = \(\frac{431}{20}\)
mean = 22

Mode:
Mode is the value which occurs the maximum number of times in a given data set.
It is the third measure of the central tendency.
16, 17, 18, 19, 19, 20, 20, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 24, 26, 26
the above data has 20 observations,
So,  24 appears most frequently, is the mode of a given data.

Median:
median : Middle value is the median of a given data set.
16, 17, 18, 19, 19, 20, 20, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 24, 26, 26
the above data has 14 observations,
Median is the average value of the 3 & 5
(22 + 22 )/2 = 44/2 = 22
So, 22 in the middle of the order sequence is the median.

Solve. Show your work. (Chapter 12)

Question 16.
The square pyramid shown has congruent triangular faces. The area of one triangular face is 48 square inches. Find the surface area of the pyramid.

Answer:
112 Sq in

Explanation:
Surface Area = Base area + Lateral Area
Base area = Side X Side
BA = 8 x 8 = 64 Sq in
Lateral  Area LA = (1/2) Side x Height
LA = (1/2) x 8 x 12
LA = 48
Surface Area SA= Base area + Lateral Area
SA = 64 + 48  = 112 Sq in

Question 17.
The length of the aquarium shown is two times its width. The height of the aquarium is 18 inches. The aquarium is filled with water to a height of 16 inches. The volume of the water is 7,200 cubic inches.
a) Find the length of the base of the aquarium.
Answer:
Length = 30 inches.

Explanation:
The height of the aquarium is 18 inches.
The aquarium is filled with water to a height of 16 inches.
The volume of the water is 7,200 cubic inches.
Volume V =Length X Width X Height
V = l.w.h
Let x be the length of the width, and length of base if 2x
V =  2w x w x 16
7200 =32 . w²
225 = w²
w = 15
width w = 15 inches
Length of the base = 2xw
l = 2 x 15
l =30 inches.

b) Then find the amount of glass, in square inches, used to make the bottom and sides of the aquarium.

Answer:
Surface Area = 2070 square inches.

Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.
SA=2lw+2lh+2hw
SA = 30×15 + 2x 18×15 + 2x 30×18
SA = 450 + 540 + 1080
SA = 2,070 sq inches.

Solve. (Chapter 13)

Question 18.
The table shows the number of hours each of 120 students spent helping their community in two months.

a) Find the value of x.
Answer:
x = 5

Explanation:
The given table shows the number of hours each of 120 students spent helping their community in two months.
x + 2x + 105 = 120
120 -105 = 3x
3x = 15
x = 5

b) Draw a histogram to represent the data. Briefly describe the data.
Answer:

Explanation:
The given table shows the number of hours each of 120 students spent helping their community in two months.
Histogram drawn using the intervals from the data in the table as shown in the above table.
Number of students is on x-axis and number of hours on y-axis.
Most of the data values are to the right of the interval 31-75,
The shape of the histogram is right skewed.

c) What percent of the students spent more than 55 hours helping their community?
Answer:
47.5 %

Explanation:
The given table shows the number of hours each of 120 students spent helping their community in two months.
The students spent more than 55 hours helping their community are,
18+16+13+10 = 57 hours
57 x 5 = 285 hours
Total 120 x 5 = 600 hours
The percent of the students spent more than 55 hours helping their community are,
57/120 = 0.475
0.475 x 100 = 47.5%

d) What percent of the students spent less than 46 hours helping their community?
Answer:
22.5%

Explanation:
The given table shows the number of hours each of 120 students spent helping their community in two months.
Number of the students spent less than 46 hours helping their community are,
5+9+13 = 27 hours.
27 x 5 = 135 hours.
Total 120 x 5 = 600 hours.
The percent of the students spent more than 55 hours helping their community are,
27/120 = 0.225
0.225 x 100 = 22.5%

Make a dot plot to show the data. Use your dot plot to answer each question. (Chapters 13, 14)

Question 19.
The table shows the results of a survey to find the number of television sets in 50 randomly chosen homes.

The total number of homes that have 0 or 1 television set is 15.

a) Find values of x and y. Then represent this set of data with a dot plot.
Answer:
x = 10
y = 4

Explanation:
The total number of homes that have 0 or 1 television set is 15.
y + 11 = 15
y = 4

total no of houses = 50
4 + 11 + 17 + x + 6 + 2 = 50
40 + x = 50
x = 10

b) Find the mean, median, and mode of the data set.
Answer:
Mean = 2.18
Median = 2
Mode = 2
Explanation:
mean = (0 x 4) + (1 x 11) + (2 x 17) + (3 x 10) + (4 x 6) + (5 x 2) ÷ 50
(11 + 34 + 30 + 24 + 10) ÷ 50
109 ÷ 50
Mean = 2.18

Median = no of houses ÷ 2 = 25
4 + 11 + 10 = 25
so the 25th home lies in the 17 homes which have 2 TV’s
Median = 2

Mode = 2
most of the houses have 2 TV’s

c) Briefly describe the data distribution and relate the measure of center to the shape of the dot plot.
Answer:
The given table shows the results of a survey to find the number of television sets in 50 randomly.
Histogram drawn using the intervals from the data in the table as shown in the above table.
Number of homes is on x-axis and number of televisions on y-axis.
Most of the data values are to the right of the interval only 1,
The shape of the histogram is right skewed.

d) A similar survey is carried out on another 30 randomly chosen homes and the mean number of television sets is found to be 1.9. If the two data sets are combined, find the mean number of television sets in the combined data set.
Answer:
Mean = 2.04

Explanation:
50 randomly chosen homes and the mean number of television sets is found to be 2.18.
30 randomly chosen homes and the mean number of television sets is found to be 1.9.
the mean number of television sets in the combined data set.
(2.18 + 1.9) ÷ 2
Mean = 2.04

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 14 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 14 Review Test Answer Key

Concepts and Skills

Solve. Show your work.

Question 1.
The data set shows nine students’ scores in a science quiz. 9, 6, 6, 5, 9, 10, 1,4, 10
Find the mean and median score.
Answer:
mean = 6\(\frac{2}{3}\)
median = 6

Explanation:
To find the mean: The sum of the values by adding them all up.
9 + 6 + 6 + 5 + 9 + 10 + 1 + 4 +10 = 60
Divide the sum by the number of values in the data set.
60 ÷ 9
Simplify as = 20 ÷ 3 = 6\(\frac{2}{3}\)
To find the median: Arrange the data points from smallest to largest.
Given set of quiz scores are {9, 6, 6, 5, 9, 10, 1,4, 10}
Ascending order of the scores are {1, 4, 5, 6, 6, 9, 9, 10, 10}
If the number of data points is odd, the median is the middle data point in the list.
So, the median is 6.

Question 2.
The mean of a set of four numbers is 3.5. If a fifth number, x, is added to the data set, the mean becomes 4. Find the value of x.
Answer:
x = 6

Explanation:
Given that,
The mean of a set of four numbers is 3.5.
Total = 3.5 x 4 = 14
If a fifth number, x, is added to the data set, the mean becomes 4.
(14 + x) ÷ 5 = 4
14 + x = 20
x = 20 – 14
x = 6

Make a dot plot to show the data. Use your dot plot to answer the question.

Question 3.
The data set shows the number of vehicles at a highway intersection during morning rush hour on 15 working days.
12, 11, 4, 6, 9, 11, 4, 6, 12, 16, 11, 10, 8, 4, 5
Find the mean, median, and mode of the data set.
Answer:
mean = 8.6
median = 9
mode = 4 and 11

Explanation:
To find the mean:
The sum of the values by adding them all up.
12 + 11 + 4 + 6 + 9 + 11 + 4 + 6 + 12 + 16 + 11 + 10 + 8 + 4 + 5 = 129
Divide the sum by the number of values in the data set.
129 ÷ 15
To find the median:
Arrange the data points from smallest to largest.
Given set of quiz scores are {12, 11, 4, 6, 9, 11, 4, 6, 12, 16, 11, 10, 8, 4, 5}
Ascending order of the scores are {4, 4, 4, 5, 6, 6, 8, 9, 10, 11, 11, 11, 12, 12, 16}
If the number of data points is odd, the median is the middle data point in the list.
So, the median is 9.
To find mode:
The mode is simply the number that appears most often within a data set.
So, 4 and 11 appears 3 times in the set.

Problem Solving

Solve. Show your work.

The data set shows the amount of money 10 children spent in a week.
$16, $13, $11, $19, $17, $28, $15, $11, $13, $11

Question 4.
Find the mean and median amount of money spent.
Answer:
mean = 15.4
median = $16

Explanation:
Mean:
The sum of the values by adding them all up.
$16 + $13 + $11 + $19 + $17 + $28 + $15 + $11 + $13 + $11 = 154
Divide the sum by the number of values in the data set.
154 ÷ 10 = 15.4
Median:
Arrange the data points from smallest to largest.
Given set of quiz scores are {$16, $13, $11, $19, $17, $28, $15, $11, $13, $11}
Ascending order of the scores are {$11, $11, $11, $13, $13, $15, $16, $17, $19, $28}
If the number of data points is even, the median is the average of middle 2 data point in the list.
$13 and $15 are the middle points in the data.
13 + 15 = 28
28 ÷ 2 = 16
So, the median is $16.

Question 5.
Which amount of money would you delete from the list if you want the mean to be closer to the median? Explain your answer.
Answer:
$28

Explanation:
Amounts at the extreme have more effect on the mean than the median.
Ascending order of the scores are {$11, $11, $11, $13, $13, $15, $16, $17, $19}

Use the data in the table to answer the question.

Question 6.
Three classes in Grade 7 took a geography test last week. The table shows the mean score of the students in each class.

The mean score of the students in classes A and B combined is 7.25. The mean score of all the students in the three classes is 6.5. Find the values of x and y.
Answer:
x = 15
y = 5

Explanation:
The mean score of the students in classes A and B combined is 7.25.
The mean score of all the students in the three classes is 6.5.
The mean = \(\frac{Sum of a set of items}{Number of items}\)
7.25 = \(\frac{6× + 25×8}{25 + x}\)
60x + 200 = 7.25{25 + x}
6x + 200 =7.25x + 181.25
200 – 181.25 = 7.25-6x
18.75 = 1.25x
x = 15

Mean
15×6 + 25×8 + 20y = 6.5(15+25+20)
90+200+20y = 6.5 x 60
290 + 20y = 390
20y = 390-290
y = 100/20
y = 5

Make a dot plot to show the data. Use your dot plot to answer questions 7 and 8.

The table shows the number of goals scored by a soccer team in 15 games.

Question 7.
Find the mean, median, and mode of the data set.
Answer:
mean = 2.2
median = 2
mode = 2

Explanation:
The above tabulated data is the table shows the number of goals scored by a soccer team in 15 games.
Mean:
The mean = \(\frac{Sum of a set of items}{Number of items}\)

mean = \(\frac{1×5 + 6×2 + 3×3 + 0×4 + 0×5 + 0×6 + 1×7}{15}\)
mean = \(\frac{5 + 12 + 9 + 7}{15}\)
mean = \(\frac{33}{15}\)
mean = 2.2
Mode
The above observations are:
1, 1, 1, 1, 1, 2, 2,2 2, 2, 2, 3, 3, 3, 7
As 2 has more number the table shows the number of goals scored by a soccer team in 15 games.
Mode is 2
Median:
The above observations are
1, 1, 1, 1, 1, 2, 2,2 2, 2, 2, 3, 3, 3, 7
total 15 observation
median : Middle value is the median of a given data set.
1, 1, 1, 1, 1, 2, 2,2 2, 2, 2, 3, 3, 3, 7
the above data has 15 observations,
2 is in the middle of the above series of numbers.
hence 2 is the median.

Question 8.
Briefly describe the data distribution and relate the measure of center to the distribution.
Answer:
The mean is to the right of the median. with mean 2.2 as peak.

Explanation:
Measures of central tendency are summary statistics that represent the center point or typical value of a dataset.
The shape of the data distribution is right skewed.
The mean gives more weight to the values on the right than the median does.
So, the mean is to the right of the median. with mean 2.2 as peak.

Use the data in the dot plot to answer questions 9 to 13.

The dot plot shows the results of a survey to find the number of computers in 30 randomly chosen families. Each dot represents 1 family.

Question 9.
What is the modal number of computers?
Answer:
2

Explanation:
The above dot plot shows the results of a survey of the number of computers in 30 randomly chosen families. Each dot represents 1 family.
8 families with 2 computes
So, the mode of the above data is 2
as 2 computers having families are more.

Question 10.
What is the mean number of computers? Round your answer to the hundredths place.
Answer:
Mean = 1.9
The nearest hundredth place is 2.

Explanation:
The mean = \(\frac{Sum of a set of items}{Number of items}\)
mean = \(\frac{0×6 + 1×6 +2×8 + 3×6 + 4×4}{30}\)
mean = \(\frac{0 + 6 +16 + 18 + 16 }{30}\)
mean = \(\frac{56}{30}\)
mean = 1.9

Question 11.
What is the median number of computers?
Answer:
median = 2

Explanation:
The above observations are:
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4
total 30 observation
median : Middle value is the median of a given data set.
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4
the above data has 30 observations,
So, 2, 2 in the middle of the order sequence is the median.
the average of (2 + 2)/2 = 4/2 = 2
hence 2 is the median.

Question 12.
Briefly describe the data distribution and relate the measure of center to the shape of the dot plot shown.
Answer:
A measure of center is called a measure of central tendency of the data plot.

Explanation:
Given, The dot plot shows the results of a survey to find the number of computers in 30 randomly chosen families. Each dot represents 1 family.

Center describes a typical value of a data point.
Two measures of center are mean and median.
The shape of the data distribution is right skewed.
Hence, the mean and median are 2.

Question 13.
A similar survey is carried out on another 15 randomly chosen families and the mean number of computers is found to be 2. If the two data sets are combined, find the mean number of computers in the combined data set. Round your answer to the nearest hundredth.
Answer:
mean = 1.86
nearest hundredth = 2

Explanation:
Number of Computers in 1st set
0×6 + 1×6 +2×8 + 3×6 + 4×4 = 56
Number of families in 1st set = 30
Number of Computers in 2nd set = x
Number of families in 2nd set = 15
Mean of 2nd set = 2
mean = \(\frac{Total number of computes in 2nd set}{Total number of families}\)
2 = \(\frac{x}{15}\)
x = 30
Total number of computers = 30 + 56 = 86
Total number of families = 15 +30 = 45

mean = \(\frac{86}{45}\)
mean = 1.91
Round the answer to the nearest hundredth.
mean 1.9