Math in Focus – Page 39 – Math in Focus Answer Key

Math in Focus Grade 1 Chapter 1 Answer Key Numbers to 10

September 20, 2024September 19, 2024 by Hillary

This handy Math in Focus Grade 1 Workbook Answer Key Chapter 1 Numbers to 10 detailed solutions for the textbook questions.

Math in Focus Grade 1 Chapter 1 Answer Key Numbers to 10

Put On Your Thinking Cap!

challenging Practice

Mother Hen’s eggs have numbers that are greater than 2 and less than 8.

Color the eggs that belong to Mother Hen.
$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 20$
Explanation:

Answer:
The numbers greater than 2 and lesser than 8 are 3, 4, 5, 6, 7,
The eggs with 4 , 6 and 7 are shaded

Put on Your Thinking Cap!

Problem Solving

$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 21$

Daryl sees a pattern made with $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ .
He wants to continue the pattern but does not know how many $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ to draw.
Draw the next group of $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ in the box below to continue the pattern.
______________
There are ___ $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ in the next group.
Answer:
In The next pattern there will be 7 triangles
1 after 3
3 after 5 and 5 after 7
$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$
There are 7 $Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 22$ in the next group.

Chapter Review/Test
Vocabulary

Match.

$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 23$
Answer:

Explanation:
The given number names are matched with the given numbers

Concepts and Skills

Circle the stars to show the number. Write the number in words.

Question 5.
$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 24$
Answer:

Explanation:
The given number is 5
so, circled the five stars and the number name of 5 is five

Question 6.
$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 25$
Answer:

Explanation:
the given number is 9
so, circled the nine stars
The number name of 9 is nine.

Fill in the blanks with greater than, less than or the same as.

$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 26$

Question 7.
The number of cups is ____ the number of saucers.
Answer:
Same
Explanation:
There are 6 cups and 6 saucers
6 = 6

Question 8.
The number of trees is ____ the number of cars.
Answer:
Greater than
Explanation:
There are 10 trees and the number of cars are 7
10 > 7

Question 9.
The number of cars is ___ the number of cups.
Answer:
Greater than
Explanation:
Number of cars are 7 and number of cups are 6
7 > 6

Write any two numbers.

Question 10.
greater than 5: ____ _____.
Answer:
6 and 7
Explanation:
5 + 1 = 6
6 + 1 = 7
The numbers greater than 5 are 6 and 7

Question 11.
less than 7: ____ _____
Answer:
6 and 5
Explanation:
7 – 1 = 6
6 – 1 = 5
The number lesser than 7 is 6 and 5

Write the missing numbers in the number pattern.

Question 12.
$Math in Focus Grade 2 Chapter 1 Answer Key Numbers to 10 27$
Answer:

Explanation:
The missing numbers are 3, 2, 1
And the given pattern is in greater number to lesser numbers.

Question 13.
4 is 1 less than ____
Answer:
5
Explanation:
5 – 1 = 4
4 is 1 less than 5

Question 14.
9 is 1 more than ______.
Answer:
8
Explanation:
9 is 1 more than 8
9 – 1 = 8

Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10

September 20, 2024September 19, 2024 by Hillary

This handy Math in Focus Grade 1 Workbook Answer Key Chapter 1 Practice 1 Counting to 10 detailed solutions for the textbook questions.

Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10

Count.

Write the numbers.

Example

$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 1$

Question 1.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 2$
Answer:

Explanation:
There are 3 lady bugs
Traced the number 3

Question 2.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 3$
Answer:

Explanation:
There are 7 pears
traced the number 7

Question 3.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 4$
Answer:

Explanation:
There are 10 cars
traced the number 10

Count.
Write the numbers.

Question 4.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 5$
Answer:

Explanation:
Counted the animals on the tree
And traced the numbers

Draw.

Question 5.
A cow has 2 horns.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 6$
Answer:

Explanation:
Drawn the horns for remaining two cows
A cow has 2 horns.
and there are 3 cows

Question 6.
A chair has 4 legs.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 7$
Answer:

Explanation:
Drawn the legs for remaining two chairs
A chair has 4 legs.
And there are 3 chairs.

Question 7.
An ant has 6 legs.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 8$
Answer:

Explanation:
Drawn the legs for remaining two ants
An ant has 6 legs.
And there are 3 ants.

Question 8.
Each ladybug has 10 spots.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 9$
Answer:

Explanation:
Each ladybug has 10 spots.
Drawn the ten dots for remaining two bugs
There are 3 bugs.

How many insects are there?
Match.

Question 9.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 10$
Answer:

Explanation:
There are 3 butterflies and matched the butterflies with the number 3
There are 6 honey bees and matched them with number 6
there are 8 spiders matched them with number 8
There are 5 wasp matched them with the number 5
there are 9 bugs matched them with the number 9

Count the things on the snowman. Circle the correct words.

$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 11$

Question 10.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 12$
Answer:

Explanation:
There are no ties for the snow man
So, circled the zero

Question 11.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 13$
Answer:

Explanation:
There are 2 scarfs for the snow man
so, circled the number two

Question 12.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 14$
Answer:

Explanation:
There are 8 circles on the snow man dress
so, circled the number eight.

Question 13.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 15$
Answer:

Explanation:
There are four birds sitting on the tree in the picture of snow man
so, circled the number four.

Question 14.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 16$
Answer:

Explanation:
The is only one that is in the mouth of snow man
so, circled the number one

Match the numbers to the words.

Question 15.
$Math in Focus Grade 1 Chapter 1 Practice 1 Answer Key Counting to 10 17$
Answer:

Explanation:
Numbers are matched with the given number names.

Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories

September 20, 2024September 19, 2024 by Hillary

This handy Math in Focus Grade 1 Workbook Answer Key Chapter 3 Practice 3 Making Addition Stories detailed solutions for the textbook questions.

Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories

Use the pictures to make addition stories.

Use number bonds to help you.

Example

$Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories 1$

Question 1.
$Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories 2$
Answer:

Explanation:
There are 5 white dolphins and 3 colored dolphins
5 + 3 = 8
The sum of 5 and 3 is 8

Question 2.
$Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories 3$
Answer:

Explanation:
cats are clapping and 4 are sleeping
6 + 4 = 10
The sum of 6 and 4 makes 10

Question 3.
$Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories 4$
Answer:

Explanation:
there are 7 girls in the competition
and 3 boys joined them
7 + 3 = 10
The sum of 7 and 3 is 10.

Question 4.
$Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories 5$
Answer:

Explanation:
Sonia has 2 star stickers
and 2 heart stickers
2 + 2 = 4
The sum of 2 and 2 is 4

Question 5.
$Math in Focus Grade 1 Chapter 3 Practice 3 Answer Key Making Addition Stories 6$
Answer:

Explanation:
There are 5 fishes which are which
and there are no colored fishes
5 + 0 = 5
In fact, the Identity Law of Addition says that any number added to zero is equal to itself.

Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds

September 20, 2024September 19, 2024 by Hillary

This handy Math in Focus Grade 1 Workbook Answer Key Chapter 2 Practice 3 Making Number Bonds detailed solutions for the textbook questions.

Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds

Match to make 8.

Question 1.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 1$
Answer:

Explanation:
Matched 2 numbers
If we add the 2 numbers the sum must be 8
2 + 6 = 8
8 + 0 = 8
4 + 4 = 8
5 + 3 = 8
7 + 1 = 8

Match the numbers.

Question 2.
Match to make 6.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 2$
Answer:

Explanation:
Matched 2 numbers
If we add the 2 numbers the sum must be 6
3 + 3 = 6
2 + 4 = 6
5 + 1 = 6
6 + 0 = 6

Question 3.
Match to make 9.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 3$
Answer:

Explanation:
Matched 2 numbers
If we add the 2 numbers the sum must be 9
8 + 1 = 9
2 + 7 = 9
3 + 6 = 9
4 + 5 = 9

Look at the picture.

Complete the number bonds.

Question 4.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 4$
Answer:

Explanation:
2 + 2 = 4
There are 4 bees and 2 flowers.
The sum of 2 and 2 is 4

Look at the number bond. Draw the correct number of butterflies.

Question 5.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 5$
Answer:

Explanation:
2 + 2 = 4
There are 2 flowers added two more flowers.
The sum of 2 and 2 is 4

Use two colors. Color the $Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 6$ to show two numbers that make the number in $Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 7$

Complete the number bonds. Fill in the blanks.

Example
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 8$

Question 6.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 9$
Answer:

Explanation:
6 + 4 = 10
The sum of 6 and 4 is 10
6 are colored in red and 4 are colored in gray.

Question 7.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 10$
Answer:

Explanation:
4 + 2 = 6
The sum of 4 and 2 is 6
There are 4 blue and 2 yellow

Question 8.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 11$
Answer:

Explanation:
6 + 2 = 8
The sum of 6 and 2 is 8
6 are colored in red and 2 are in green

Use two colors.
Color the $Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 12$ to show two numbers that make 5. Complete the number bonds. Fill in the blanks.

Question 9.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 13$
Answer:

Explanation:
3 + 2 = 5
The sum of 3 and 2 makes 5
3 are colored in red and 2 are colored in blue.

Question 10.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 14$
Answer:

Explanation:
4 + 1 = 5
The sum of 4 and 1 is 5
4 are colored in red and 1 is colored in orange

Question 11.
$Math in Focus Grade 1 Chapter 2 Practice 3 Answer Key Making Number Bonds 15$
Answer:

Explanation:
2 + 3 = 5
The sum of 2 and 3 is 5
2 are colored white and 3 are colored brown.

Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add

September 20, 2024September 19, 2024 by Hillary

This handy Math in Focus Grade 1 Workbook Answer Key Chapter 3 Practice 2 Ways to Add detailed solutions for the textbook questions.

Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add

Complete the number bonds. Then fill in the blanks.

Example

$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 1$

Question 1.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 2$
Answer:

Explanation:
8 + 2 = 10
2 + 8 = 10
The combination of numbers gives the same sum
to make a number bond of 10
used 8 and 2 numbers and made different combinations.

Question 2.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 3$
Answer:

Explanation:

Question 3.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 4$
Answer:

Explanation:

Complete the number bonds. Then fill in the blanks.

Question 4.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 5$
Answer:

Explanation:

Question 5.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 6$
Answer:

Explanation:

Question 6.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 7$
Answer:

Explanation:

Question 7.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 8$
Answer:

Explanation:

Question 8.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 9$
Answer:

Explanation:

Question 9.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 10$
Answer:

Explanation:

Help each Momma Butterfly find her babies! Color the small butterflies that match her number.

Question 10.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 11$
Answer:

Explanation:

Question 11.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 12$
Answer:

Explanation:

Question 12.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 13$
Answer:

Explanation:

Question 13.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 14$
Answer:

Explanation:

Question 14.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 15$
Answer:

Explanation:

Add.
You can draw number bonds to help you.

Question 15.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 16$
Answer:

Explanation:

Now color the train cars above. Then fill in the table with your answers.

Question 16.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 17$
Answer:

Explanation:

Solve.

Question 17.
A ball falls into the number machine. Which ball is it? Write the correct number on the ball below.
$Math in Focus Grade 1 Chapter 3 Practice 2 Answer Key Ways to Add 18$
Answer:
Explanation:

Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping

September 20, 2024September 18, 2024 by Hillary

Go through the Math in Focus Grade 2 Workbook Answer Key Chapter 2 Practice 1 Addition Without Regrouping to finish your assignments.

Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping

Add.

Question 1.
232 + 645 = ?
Add the ones.
2 ones + 5 ones = 7 ones
Add the tens.
3 tens + 4 tens = _________ tens
Add the hundreds.
2 hundreds + 6 hundreds = _______ hundreds
232 ÷ 645 = _______
$Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping 1$
Answer:

Add the ones.
2 ones + 5 ones = 7 ones
Add the tens.
3 tens + 4 tens = 7 tens
Add the hundreds.
2 hundreds + 6 hundreds = 8 hundreds

Question 2.
$Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping 2$
Answer:

Explanation:
Add the ones.
8 ones + 1 ones = 9 ones
Add the tens.
0 tens + 7 tens = 7 tens
Add the hundreds.
1 hundreds + 8 hundreds = 9 hundreds

Question 3.
$Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping 3$
Answer:

Explanation:
Add the ones.
9 ones + 0 ones = 9 ones
Add the tens.
7 tens + 2 tens = 9 tens
Add the hundreds.
8 hundreds = 8 hundreds

Question 4.
122 + 473 = ___
$Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping 4$
Answer:

Explanation:
The sum of 122 + 473 = 595

Question 5.
217 + 771 = ___
$Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping 5$
Answer:

Explanation:
The sum of 217 + 771 = 988

Add.

Question 6.
Three friends have some addition cards. Some answers will win prizes. Help them add to find their prizes.
$Math in Focus Grade 2 Chapter 2 Practice 1 Answer Key Addition Without Regrouping 6$
Answer:

Explanation:
First friend has bear
second friend has book
third friend has hat

Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000

September 20, 2024September 18, 2024 by Hillary

Go through the Math in Focus Grade 2 Workbook Answer Key Chapter 2 Addition up to 1,000 to finish your assignments.

Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000

Put on Your Thinking cap!

challenging practice

Write the missing numbers.

Question 1.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 1$
Answer:

Explanation:
In ones place 3 and 1 makes 4
so, missing number is 3

Question 2.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 2$
Answer:

Explanation:
In hundreds place 6 hundreds and 2 hundreds make 8 hundreds

Question 3.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 3$
Answer:

Explanation:
In tens place 0 tens + 9 tens is 9 tens

Question 4.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 4$
Answer:

Explanation:
In tens place 2 tens and 4 tens make 6 tens

Question 5.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 5$
Answer:

Explanation:
7 ones + 5 ones = 12
1 tens and 2 ones
In tens place 3 tens 9 tens and 1 ten = 13 tens

Question 6.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 6$
Answer:

Explanation:
In ones place 3 ones and 9 ones make 12 ones
That is 1 tens and 2 ones
1 ten will be carry
So, 8 ten +5 ten + 1 ten ten = 8 + 6 = 14 tens

Do these.

Question 7.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 7$
Answer:

Explanation:
The numbers are written in least to greatest

a. Order the numbers from greatest to least.
Answer:

Explanation:
The number are written in Decending order

b. Add 100 to the least number. Show your work.
Answer: 554
Explanation:
The least number in the given numbers is 454 + 100 = 554

Put On Your Thinking Cap!

problem Solving

Make two 3-digit numbers from the numbers below. Use each number only once. What are the two 3-digit numbers that give the greatest answer when you add them?

$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 8$

____ ____ _____
Answer:

Explanation:
5, 1, 2 = 5 + 1 + 2 = 8
and 4, 3, 0 = 4 + 3 + 0 = 7
are the two 3-digit numbers that give the greatest answer when you add them

Chapter Review/Test

Vocabulary

Fill in the blanks with words from the box.
The words may be used more than once.

Question 1.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 9$
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 10$
Step 1
Add the _____
2 ones + 8 ones = 10 ones
__________ the ones.
10 ones = 1 ten 0 ones

Explanation:
Add the ones
2 ones + 8 ones = 10 ones
Regroup the ones.
10 ones = 1 ten 0 ones
Step 2
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 11$
Add the ________
1 ten + 6 tens + 6 tens = 13 tens
___________ the tens.
13 tens = 1 _________ 3 ____

Explanation:
Add the tens
1 ten + 6 tens + 6 tens = 13 tens
Regroup the tens.
13 tens = 1tens 3 ones
Step 3
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 12$
Add the ____
1 hundred + 4 hundreds + 2 hundreds
= 7 hundreds
462 + 268 = 730

Explanation:
Add the hundreds
1 hundred + 4 hundreds + 2 hundreds
= 7 hundreds
462 + 268 = 730

Concepts and Skills

Add.
Then match the problems with the same answer.

Question 2.
$Math in Focus Grade 2 Chapter 2 Answer Key Addition up to 1,000 13$
Answer:

Explanation:
The addition of three digits number is found on both sides
And matched the sum of numbers

Problem Solving

Solve.

Question 3.
Mr. Thomas drives 173 miles on Monday. On Tuesday, he drives 216 miles. How many miles does he drive in all?
He drives ___ miles in all.
Answer:
He drives ___ miles in all.
Explanation:
Mr. Thomas drives 173 miles on Monday.
On Tuesday, he drives 216 miles.
173 + 216 = 389 miles does he drive in all

Question 4.
A carpenter has 362 pieces of lumber. He needs another 228 pieces of lumber to build a bridge. How many pieces of lumber does he need to build the bridge?
He needs ___ pieces of lumber to build the bridge.
Answer:
He needs ___ pieces of lumber to build the bridge
Explanation:
A carpenter has 362 pieces of lumber.
He needs another 228 pieces of lumber to build a bridge.
362 + 228 = 590 pieces of lumber does he need to build the bridge

Solve.

Question 5.
A movie theater sells 294- tickets to the first show. It sells 457 tickets to the second show. How many tickets does it sell in all?
It sells ___ tickets in all.
Answer:
It sells ___ tickets in all
Explanation:
A movie theater sells 294- tickets to the first show.
It sells 457 tickets to the second show.
294 + 457 = 751 tickets does it sell in all

Question 6.
Shantel has 546 stickers in her collection. She has 278 fewer stickers than Sherice. How many stickers does Sherice have in her collection?
Shantel has ___ stickers in her collection.
Answer:
Shantel has ___ stickers in her collection
Explanation:
Shantel has 546 stickers in her collection.
She has 278 fewer stickers than Sherice.
546 + 278 = 824 stickers does Sherice have in her collection

Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key

September 20, 2024September 18, 2024 by Hillary

This handy Math in Focus Grade 8 Workbook Answer Key Cumulative Review Chapters 10-11 detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key

Construct a scatter plot for each table of bivariate data. Identify any outliers. (Lesson 10.1)

Question 1.
Use 1 centimeter on the horizontal axis to represent 1 unit and 1 centimeter on the vertical axis to represent 100 units.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 1$
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_1$

Question 2.
Use 1 centimeter on the horizontal axis to represent 0.2 units for x interval from 36.0 to 38.0 and 1 centimeter on the vertical axis to represent 2 units for y interval from 230 to 252.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 2$
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_2$

Describe any association between the bivariate data in each scatter plot. (Lesson 10.1)

Question 3.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 3$
Answer:
One variable increasing with another has a positive association, otherwise, it is negative. If the scatter plot follows a straight line then it is the linear association, otherwise, it is nonlinear association. If it is hard to find the arrangement of the variables then it has a weak association, otherwise, it has a strong association. In this case, the graph has positive, strong and linear associations.

Question 4.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 4$
Answer:
If it is hard to identify the arrangement of the variables then it has a weak association and when it seems it has no pattern at all then it has no association. For this case, the graph has no association at all.

State the line that represents a line of best fit for each scatter plot. (Lesson 10.2)

Question 5.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 5$
Answer:
Line B fits the scatter plot, as it is the straight line that shows the data on scatter plot when compared with the Line A.

Question 6.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 6$
Answer:
Line C fits the scatter plot, It is the most possible straight line that shows the data on a scatter plot in comparison with Line A and Line B.

Construct each scatter plot and draw a line of best fit for the given table of bivariate data. (Lesson 10.2)

Question 7.
Use 1 centimeter on the horizontal axis to represent 1 unit and 1 centimeter on the vertical axis to represent 2 units.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 7$
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_7$

Question 8.
Use 1 centimeter to represent 1 unit on both axes.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 8$
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_8$

Identify whether the given data is qualitative or quantitative. (Lesson 10.3)

Question 9.
Number of eggs hatched
Answer: Quantitative

Question 10.
Ice cream flavors
Answer: Qualitative

Copy and fill in the missing information in the two-way table. (Lesson 10.3)

Question 11.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 9$
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_11$

Construct a two-way table using the given data. (Lesson 10.3)

Question 12.
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 10$
W represents watch the documentary
NW represents did not watch the documentary
P represents passed the science test
NP represents did not pass the science test
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_12$

Tell whether the outcomes described are from a simple or compound event. If it is a compound event, identify the simple events that form the compound event. (Lesson 11.1)

Question 13.
Drawing a blue marble followed by a green marble, without replacing the first marble, from a box containing 2 red, 5 blue, and 3 green marbles.
Answer:
Compound event is composed of more than 1 simple event, Since there are two draws, the first draw to obtain a blue marble and the second one is to pick the green marble. Therefore, this is a compound event.

Question 14.
Getting a number between 2 and 5 when rolling a fair six-sided number die.
Answer:
Rolling a die to obtain a number between 2 and 5 once is one single event. It is possible to get in one single event, therefore this is a simple event.

Question 15.
Getting a product of 12 when rolling two fair six-sided number dice.
Answer:
Since there are two dices to roll, the first roll is the first event and the second dice is rolled to get 12 when the digit obtained is multiplied is the second event. Therefore, this is a compound event.

Solve. Show your work.

Question 16.
A fair four-sided number die, labeled 4 to 7, and a letter cube, labeled A, B, C, D, E, and E, are rolled. (Lessons 11.1, 11.2)
a) Draw a possibility diagram to represent the possible outcomes.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_16$

b) Find the probability of getting the letter E.
Answer:
There are 24 possible outcomes and 8 have E,
Probability = favorable outcomes / total outcomes
= 8/24
= 1/3

Question 17.
Two fair six-sided number dice numbered 1 to 6 are rolled. (Lessons 11.1, 11.2)
a) Draw a possibility diagram to represent the possible outcomes.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_17$

b) Find the probability that the outcomes for both six-sided number dice are even.
Answer:
There are total of 36 possibe outcomes and 9 obtained even numbers on both the dice
Probability = favorable outcomes/total outcomes
= 9/36
= 1/4

c) Find the probability that the product of the two numbers is a prime number.
Answer:
There are total of 36 possibe outcomes and 6 obtained prime numbers on both the dice
Probability = favorable outcomes/total outcomes
= 6/36
= 1/6

Question 18.
Spinner P is divided into 5 equal areas and labeled from 5 to 9. The spinner is spun and a coin is tossed. (Lessons 11.1, 11.2)
a) Draw a possibility diagram to represent the possible outcomes.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_18a$

b) Find the probability of getting heads and an even number.
Answer:
There are total of 10 possibe outcomes and 2 results have head and even numbers in it
Probability = favorable outcomes/total outcomes
= 2/10
= 1/5

Question 19.
There are 4 yellow highlighters, 2 blue highlighters, and 3 purple highlighters in bag A. There are 2 white balls and 4 blue balls in bag B. A highlighter is randomly drawn from bag A, and then a ball is randomly drawn from bag B. (Lesson 11.3)
a) Draw a tree diagram to represent the possible outcomes and their corresponding probabilities.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_19a$

b) What is the probability of drawing a yellow highlighter and a blue ball?
Answer:
For the probability of obtaining a blue ball after obtaining a yellow highlighter denoted as P(Y,B), the probability of getting a yellow highlighter in the Bag A is 4/9 while the probability of selecting a blue ball in the Bag B is P(B) after getting a yellow highlighter is 4/6 then,
Probability = P(Y) x P(B)
= 4/9 x 4/6
= 16/54
= 8/27

c) What is the probability of drawing a highlighter and a ball that are the same color?
Answer:
For the probability of obtaining a blue highlighter followed by a blue ball denoted as P(B,B), the probability of getting a blue highlighter in the Bag A P(B) is 2/9 while the probability of selecting a blue ball in the Bag B is P(B) is 4/6 then,
Probability = P(Y) x P(B)
= 2/9 x 4/6
= 8/54
= 4/27

d) What is the probability of drawing exactly one blue item?
Answer:
The probability of selecting one blue item is the probability of getting one purple highlighter and a blue ball P(P,B) or the probability of selecting yellow highlighter and a blue ball P(Y,B) or the probability of obtaining a blue highlighter and white ball P(B,W)
P(P,B)+P(Y,B)+P(B,W)
= P(P) . P(B) + P(Y) . P(B) + P(B) . P(W)
= 3/9 x 4/6 + 4/9 x 4/6 + 2/9 x 2/6
= 12/54 + 16/54 + 4/54
= 32/54
= 16/27

Question 20.
A box contains 16 color cards: 5 red, 4 blue, and 7 green. Two cards are picked at random from the box, one after another without replacement. The tree diagram shows the possible outcomes and the corresponding probabilities. (Lesson 11.4)
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 11$
a) Find the values of p, q, r, s, and t.
Answer:
To find the value of p:
P(B) + P(R) + p = 1
4/16 + 5/16 + p = 1
9/16 + p = 1
p = 1 – 9/16
p = 7/16
To find the value of q:
P(G) + P(R) +p = 1
7/15 + 4/15 + p = 1
11/15 + p = 1
p = 1 – 11/15
p = 4/15
To find the value of r:
The probability of obtaining red in the second pick is 5/15 or 1/3.
To find the value of s:
The probability of obtaining blue denoted as s in the second pick is 4/15.
To find the value of t:
The probability of obtaining green denoted as t in the second pick is 6/15.

b) Find the probability that both cards are green.
Answer:
P(G,G) = P(G) x P(G)
= 7/16 x 6/15
= 42/240
= 7/40

c) Find the probability that both cards are the same color.
Answer:
P(G,G) + P(B,B) + P(R,R)
=P(G) x P(G) + P(B) x P(B) + P(R) x P(R)
=7/16 x 6/15 + 4/16 x 3/15 + 5/16 x 4/15
= 42/240 + 12/240 + 20/240
= 74/240
= 37/120

d) Find the probability that both cards are different colors.
Answer:
P(M) + P(NM) = 1
37/120 + P(NM) = 1
P(NM) = 1 – 37/120
P(NM) = 83/120

e) Find the probability that both cards are blue or both are green.
Answer:
P(G,G) or P(B,B) = P(G,G) + P(B,B)
= P(G) x P(G) + P(B) x P(B)
= 7/16 x 6/15 + 4/16 x 4/15
= 42/240 + 12/240
= 54/240
= 9/40

Problem Solving

Solve. Show your work.

Question 21.
Alan has 3 pencils in a box: 1 blue, 1 red, and 1 orange. He has 1 blue marker and 2 red markers in another box. Alan randomly selects a pencil and a marker. Use a possibility diagram to represent the possible outcomes. Then find the probability of getting the same colors for the pencil and marker. (Chapter 11)
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_21$
There are total of 9 outcomes and 3 favorable outcomes, then
Probability = favorable outcomes / total outcomes
= 3/9
= 1/3

Question 22.
Joyce rolled a fair six-sided number die labeled 1 to 6 and a fair four-sided number die labeled 2 to 5. Use a possibility diagram to find the probability that the product of the two resulting numbers is at least 20. (Chapter 11)
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_22$
Explanation:
There are 36 total outcomes and 5 are greater than or equal to 20 then,
Probability = favorable outcomes / total outcomes
= 5/36

Refer to the scenario below to answer questions 23 to 28.

The data below shows the amount of electricity consumption and cost for some households in a particular month. (Chapter 10)
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 12$
Question 23.
Construct a scatter plot for the given bivariate data. Use 1 centimeter on the horizontal axis to represent 20 kilowatt-hour for the x interval from 800 to 1,000 and 1 centimeter on the vertical axis to represent 2 dollars for the y interval from 86 to 108.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_23$

Question 24.
Describe the association between the electricity consumption and the cost.
Answer: The variable have a positive and linear association. Then, a high electricity usage corresponds to a high cost and a low consumption means a lower price.

Question 25.
Draw a line of best fit.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_25$

Question 26.
Write an equation for the line of best fit. Use points (820, 88) and (900, 97).
Answer:
For the equation of the line that best fits , using the points that best fits line passes through (820,88) and (900,97) to solve the slope of the line.
Slope of the line m = y2 – y1/ x2 – x1
m = 97 – 88 / 900 – 820
m = 9 / 80
m = 0.11
Using the slope intercept form given by y = mx + b, where m is the slope, b is the y intercept
y = mx + b
88 = (0.11)820 + b
88 = 90.2 + b
b = 90.2 – 88
b = 2.2
Therefore, the equation of line best fits is y = 0.11x + 2.2

Question 27.
Using the equation in 26, predict the cost per month for an electricity consumption of 888 kilowatt hour in that particular month.
Answer:
To estimate the price x given the electricity usage y, study the graph and the line of best fit at the point y at 888 kw. The point is on (888, 96.8).
Therefore, a household with 888 kw per hour consumption will pay around $96.8

Question 28.
Using the equation in 26, predict the cost per month for an electricity consumption of 1,000 kilowatt hour in that particular month.
Answer:
By using the equation of the line the cost of electricity consumption of 1000 kw per hour is calculated as
y = mx + b
y = 0.11(1000) + 2.2
y = 110 + 2.2
y = 112.2

Solve. Show your work.

Question 29.
Bag A contains 25 tomatoes, 4 of which are rotten. Bag B contains 15 tomatoes, 6 of which are rotten. Jamie randomly selects a tomato from bag A followed by another random selection from bag B. (Chapter 11)
a) Draw a tree diagram to represent the possible outcomes and their corresponding probabilities.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_29a$

b) What is the probability that both tomatoes are rotten?
Answer:
P(R,R) = P(R) x P (R)
= 4/25 x 6/15
=24/375
= 8/125

c) What is the probability that exactly one tomato is rotten?
Answer:
P(NR,R) or P(R,NR) = P(NR,R) + P(R,NR)
= P(NR) x P(R) + P(R) x P(NR)
= 21/25 x 6/15 + 4/25 x 9/15
= 126/375 + 36/375
= 162/375
= 54/375

d) What is the probability that at least one tomato is not rotten?
Answer:
P(R,NR) + P(NR,R) + P(R,R)
= P(R) x P(NR) + P(NR) x P(R) + P(R) x P(R)
= 4/25 x 9/15 + 21/25 x 6/15 + 4/25 x 6/15
= 36/375 + 126/375 + 24/375
= 186/375
= 62/125

Question 30.
The probability that Jeremy wakes up late is 0.3. Jeremy can choose to cycle, take a bus, or drive a car to the gym. If he does not wake up late, the probability that he cycles is 0.5, takes a bus is 0.4, and drives a car is 0.1. If he wakes up late, the probability that he cycles is 0.05, takes a bus is 0.2, and drives a car is 0.75. (Chapter 11)
a) Draw a tree diagram to represent the possible outcomes and their corresponding probabilities.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_30a$

b) Find the probability that Jeremy will wake up and cycle to the gym.
Answer:
P(L,BC) = P(L) x P(BC)
= 0.3 x 0.05
= 0.015

c) Find the probability that Jeremy will not wake up late and will drive a car to the gym.
Answer:
P(E,C) = P(E) x P(C)
= 0.7 x 0.1
= 0.07

d) Find the probability that Jeremy will take a bus to the gym.
Answer:
P(E,B) or P(L,B) = P(E,B) + P(L,B)
= P(E) x P(B) + P(L) x P(B)
= 0.7 x 0.4 + 0.3 x 0.2
= 0.28 + 0.06
= 0.34

Refer to the scenario below to answer questions 31 to 34.

The data below shows the gender and favorite sport of 12 students surveyed. (Chapter 10)
$Math in Focus Grade 8 Cumulative Review Chapters 10-11 Answer Key 13$
Question 31.
Construct a two-way table using the above data.
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_31$

Question 32.
Describe the association between genders and favorite sports based on the given data.
Answer:
The total number of female who answers the survey is greater than the number of boys who accomplished the same activity. Swimming got the most number of votes compared to the other two. Therefore, more female students engage to sports compared to the male population and the most favorite sport is swimming.

Question 33.
Copy and find the relative frequencies (to the nearest hundredth) to compare the distribution of the genders among the different favorite sports (columns).
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_33$

Question 34.
Copy and find the relative frequencies (to the nearest hundredth) to compare the distribution of the favorite sports among the genders (rows).
Answer:
$Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 10-11 Answer Key_34$

Math in Focus Grade 8 Chapter 11 Review Test Answer Key

September 20, 2024September 18, 2024 by Hillary

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key

Concepts and Skills

State whether each event is a simple or compound event.

Question 1.
Drawing 2 yellow marbles ¡n a row from a bag of yellow and green marbles.
Answer:
Drawing 2 yellow marbles in a row from a bag of yellow and green marbles are compound events. Because the number of marbles is 3 so, the result is the number of outcomes.

Question 2.
Drawing 1 red pebble and 1 yellow pebble in a row from a bag of red and yellow pebbles.
Answer:
Drawing 1 red pebble and 1 yellow pebble in a row from a bag of red and yellow pebbles is compound. Because the number of pebbles is 2 so, the result is the number of outcomes

Question 3.
Tossing a coin once.
Answer:
Tossing a coin once is a simple event because it has a result of one outcome.

Draw the possibility diagram and state the number of possible outcomes for each compound event.

Question 4.
From three cards labeled A, B, and C, draw two cards, one at a time with replacement.
Answer:
$Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key_4$

Question 5.
From a pencil case with 1 red pen, 1 green pen, and 1 blue pen, select two pens, one at a time without replacement.
Answer:
$Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key_5$

Question 6.
Toss a fair four-sided number die, labeled 1 to 4, and a coin.
Answer:
$Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key_6$

Draw the tree diagram for each compound event.

Question 7.
Spinning a spinner divided into 4 equal areas labeled 1 to 4, and tossing a coin.
Answer:
$Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key_6$

Question 8.
Picking two green apples randomly from a basket of red and green apples.
Answer:
The probability of picking red and green apples is 1/2.

State whether each compound event consists of independent events or dependent events.

Question 9.
From a pencil case, two-color pencils are randomly drawn, one at a time without replacement.
Answer: Dependent event

Question 10.
From two classes of 30 students, one student ¡s selected randomly from each class for a survey.
Answer: Independent event

Problem Solving

Solve. Show your work.

Question 11.
There are two tables in a room. There are 2 history textbooks and 1 math textbook on the first table. There are 1 history workbook and 1 math workbook on the second table. Use a possibility diagram to find the probability of randomly selecting a history textbook from the first table and a math workbook from the second table.
Answer:
Given,
There are two tables in a room. There are 2 history textbooks and 1 math textbook on the first table.
1 + 2 = 3
The probability of randomly selecting a history textbook from the first table and a math workbook from the second table is 1/3.

Question 12.
A fair four-sided number die is marked 1, 2, 2, and 3. A spinner equally divided into 3 sectors is marked 3, 4, and 7. Jamie tosses the number die and spins the spinner.
a) Use a possibility diagram to find the probability that the sum of the two resulting numbers is greater than 5.
Answer:
Note that there are 4 possible outcomes in rolling a die which is 1, 2, 2, and 3, and 3 outcomes for the spinner which are 3, 4 and 7. Add these possible resuLts, then mark the answers that are bigger the 5.
$Math in Focus Grade 8 Chapter 11 Review Test Answer Key 2$
Observe that there are 12 possible outcomes and 8 are greater than 5 then
$Math in Focus Grade 8 Chapter 11 Review Test Answer Key 3$
Therefore, the probability of obtaining a number that is larger than 5 is $\frac{2}{3}$.

b) Use a possibility diagram to find the probability that the product of the two resulting numbers is odd.
Answer:
Notice that there are 4 possible outcomes in rolling a die which is 1, 2, 2. and 3, and 3 outcomes for the spinner which are 3, 4 and 7. Multiply these possible results, then mark the answers that odd numbers.
$Math in Focus Grade 8 Chapter 11 Review Test Answer Key 4$
Observe that there are 12 possibLe outcomes and 4 are odd numbers then
$Math in Focus Grade 8 Chapter 11 Review Test Answer Key 5$
Therefore, the probability of obtaining an odd number is $\frac{1}{3}$.

Question 13.
A juggler is giving a performance by juggling a red ball, a yellow ball, and a green ball. All 3 balls have equal chance of dropping. If one ball drops, the juggler will stop and pick up the ball and resume juggling. If another ball drops again, the juggler will stop the performance.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:
$Math-in-Focus-Grade-8-Course-3-B-Chapter-11-Review-Test-Answer-Key_13a$

b) Find the probability of dropping the same colored ball twice.
Answer: 1/3

c) Find the probability of dropping one green and one yellow ball.
Answer:
P(G)P(R) = 1/3 × 1/3= 1/9

Question 14.
In a marathon, there is a half-marathon and a full-marathon. There are 60 students who participated in the half-marathon and 80 participated in the full-marathon. Half of the students in the half-marathon warm up before the run, while three-quarters of the students in the full-marathon warm up. Assume that warming up and not warming up are mutually exclusive and complementary.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:
Since, there are 2 types of marathon, then start the tree diagram with 2 branches. Now, at the end of these branches put the possible outcomes: H for half-marathon, and F for full marathon. Note that the probability of full marathon is $\frac{80}{140}$ and the probability of half-marathon is $\frac{60}{140}$. Next, there are a group who warms up and who does not, then make another 2 branches. At the end of these branches put the possible outcomes: W for warming up and NW for not warming up. Finally, conclude the possible results. Therefore, the tree diagram for the said event is as depicted below.
$Math in Focus Grade 8 Chapter 11 Review Test Answer Key 6$

b) What is the probability of randomly picking a marathon participant who warms up before running a full-marathon?
Answer: full marathon: 80 runners

c) What is the probability of randomly picking a marathon participant who does not warm up before running?
Answer:
80 + 60 = 140
total: 140 runners; 90 warm-up.
The probability that a randomly selected runner warms up is 90/140.

Question 15.
The probability of Cindy waking up after 8 A.M. on a weekend day is p. Assume the events of Cindy waking up after 8 A.M. and by 8 A.M. are mutually exclusive and complementary.
a) If p = 0.3, find the probability that she will wake up after 8 A.M. on two consecutive weekend days.
Answer:
P(waking up after 8 A.M. two weekend days in a row) = p²
If p=0.3
p² = (0.3)²
p² = 0.09

b) If p = 0.56, find the probability that she will wake up by 8 A.M. on two consecutive weekend days.
Answer:
P(waking up before 8 A.M two weekend days in a row) = (1 – p)²
If p = 0.56, 1-p = 0.44 and
(1 – p)² = (0.44)² = 0.1936.

Question 16.
In a jar, there are 2 raisin cookies and 3 oat cookies. Steven takes two cookies one after another without replacement.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:
Since there are two different kinds of muffins, which are bran and pumpkin, then start the tree diagram with 2 branches. Now, at the end of these branches put the possible outcomes: P for pumpkin muffin, and B for bran muffins. Note that there are 2 pumpkin muffins out of the total of 5 muffins, and 3 bran muffins out of the 5 total muffins. So the probability for selecting a pumpkin is $\frac{2}{5}$, and the probability of selecting a bran is $\frac{3}{5}$.

Next, there are stilt 2 different kinds of muffins, then again, at the end of these branches put the possible outcomes: P for pumpkin muffin, and B for bran muffins. If a pumpkin muffin is obtained in the first pick, then on the second pick, there are now 1 pumpkin muffin and a totat of 4 muffins left since a muffin is already taken, then the probability of choosing a pumpkin is $\frac{1}{4}$. Also, there are still 3 bran muffin out of 4 muffins left, thus, the probability of selecting a bran muffin is $\frac{3}{4}$.

If a bran muffin is obtained in the first pick, then on the second pick, there are still a pumpkin muffin and a total of 4 muffins left since a muffin is already taken, then the probability of choosing a pumpkin is $\frac{2}{4}$. Also, there are only 2 bran muffin left out of 4 muffins, thus, the probability of selecting a bran muffin is $\frac{2}{4}$. Therefore, the tree diagram for the said event is as depicted below.
$Math in Focus Grade 8 Chapter 11 Review Test Answer Key 7$

b) Find the probability of Steven randomly getting two of the same type of cookie.
Answer:
The probability of seLecting 2 muffins of a matching type is the probability of obtaining 2 pumpkin muffins, P(P, P), or the probability of choosing 2 bran muffins, P(B, B), then
P(P, P) or P(B, B) = P(P, P) + (B, B)
= P(P) · P(P) +P(B) · P(B)
= $\frac{2}{5}$ · $\frac{1}{4}$ + $\frac{3}{5}$ · $\frac{2}{4}$
= $\frac{2}{20}$ + $\frac{6}{20}$
= $\frac{8}{20}$
Therefore, the probability of picking 2 muffins of a matching type is $\frac{8}{20}$.

c) Find the probability of Steven randomly getting at least one raisin cookie.
Answer:
For, the probability of selecting a minimum of one pumpkin muffin, note that the probability of choosing 2 bran muffins, P(B, B), and the probability of picking at least one pumpkin muffin is complementary.
P(B, B) + P(Minimum of One Pumpkin) = 1
P(Minimiun of One Pumpkin) = 1 – P(B, B)
P(Minimum of One Pumpkin) = 1 – P(B) · P(B)
P(Minimum of One Pumpkin) = 1 – $\frac{3}{5}$ · $\frac{2}{4}$
P(minimum of One Pumpkin) = 1 – $\frac{6}{20}$
P(Minimum of One Pumpkin) = $\frac{7}{10}$
Therefore, the probabiUty of picking a minimum of one pumpkin is $\frac{7}{10}$.

Question 17.
Out of 100 raffle tickets, 4 are marked with a prize. Matthew randomly selects two tickets from the box.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:
$Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key_17a$

b) What is the probability that Matthew does not win any prizes?
Answer: 152/165

c) What is the probability that Matthew gets exactly one of the prizes?
Answer: 64/825

Question 18.
The tree diagram shows the probability of how Shane spends his day gaming or cycling, depending on the weather. The probability of rain is denoted by a. Assume that gaming and cycling are mutually exclusive.
$Math in Focus Grade 8 Chapter 11 Review Test Answer Key 1$
a) If a = 0.4, find the probability that he will spend his day gaming.
Answer:
Given that a = 0.4
Here the G represents gaming.
The probability that he will spend his day on gaming =(1- 0.4) × 1/4 = 0.15.

b) If a = 0.75, find the probability that he will spend his day cycling.
Answer:
Given that a = 0.75
Here C represents cycling.
The probability that he will spend his day in cycling = (1-0.75) × 3/4 = 0.1875.

Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events

September 20, 2024September 18, 2024 by Hillary

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.4 Dependent Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.4 Answer Key Dependent Events

Math in Focus Grade 8 Chapter 11 Lesson 11.4 Guided Practice Answer Key

Solve. Show your work.

Question 1.
A deck of four cards with the letters D, E, E, D are placed facing down on a table. Two cards are turned at random to show the letter. Draw a tree diagram to represent the possible outcomes in this compound event.
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 1$
Let D represent letter D and E represent letter E.
1st draw
P(D) = $\frac{2}{4}$
P(E) = $\frac{2}{4}$

2nd draw
P(D after D)4 = $\frac{?}{?}$ There is $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ D left after 1 D is drawn.
P(E after D) = $\frac{?}{?}$ There are $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ E still after 1 D is drawn.
P(D after E) = $\frac{?}{?}$ There are $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ D still after 1 E is drawn.
P(E after E) = $\frac{?}{?}$ There is $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ E left after 1 E is drawn.
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 3$
Answer:
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Guided Practice Answer Key_1$
P(D after D)4 = There is 1/3 D left after 1 D is drawn.
P(E after D) = There are 2/3 E still after 1 D is drawn.
P(D after E) = There are 2/3 D still after 1 E is drawn.
P(E after E) = There is 1/3 E left after 1 E is drawn

Question 2.
There are 16 different color pebbles in a jar. 11 of them are blue and the rest are orange. Two pebbles are randomly selected from the jar, one at a time without replacement.
a) Find the probability of taking an orange pebble followed by a blue pebble.
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 4$
The probability of randomly taking an orange pebble followed by a blue pebble is $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$
Answer:
P(B,O) = P(B) x P(O)
= 11/16 x 5/15
= 55/240
= 11/48
~0.23
Therefore, the probability of selecting a blue than a orange pebble is 11/48 or approx 0.23.

b) Find the probability of taking two orange pebbles.
P(O, O) = P(O) • P(O after O)
= $\frac{?}{?} \cdot \frac{?}{?}$
= $\frac{?}{?}$
The probability of randomly taking two orange pebbles is $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ .
Answer:
P(O, O) = P(O) • P(O)
= 5/16 x 4/15
= 20/240
= 1/12
~0.08
Therefore, the probability of selecting two orange pebbles is 1/12 or approx 0.08.

c) Find the probability of taking two blue pebbles.
P(B, B) = P(B) • P(B after 8)
= $\frac{?}{?} \cdot \frac{?}{?}$
= $\frac{?}{?}$
The probability of randomly taking two blue pebbles is $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ .
Answer:
P(B, B) = P(B) • P(B)
= 11/16 x 10/15
= 110/240
= 11/24
~0.46
The probability of randomly taking two blue pebbles is 11/24 or approx 0.46.

Question 3.
The tree diagram below shows how passing an examination depends on whether a student studies (S) or does not study (NS) for the exam. The probability that a student studies is denoted by p. Assume that S and NS are mutually exclusive events.
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 5$
a) If p = 0.4, find the probability that a student passes the examination.
P(S) = $\frac{?}{?}$ Write the fraction for 0.4.
P(NS) = 1 – P(S) Events S and NS are complementary.
= 1 – $\frac{?}{?}$
= $\frac{?}{?}$
P(P) = $\frac{?}{?}$ • $\frac{?}{?}$ + $\frac{?}{?}$ • $\frac{?}{?}$ Evaluate P(S, P) + P(NS, P).
= $\frac{?}{?}$
If the probability of studying is 0.4, then the probability of passing is $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ .
Answer:
P(NS) + P(S) = 1
P(NS) = 1 – P(S)
P(NS) = 1 – 4/10
P(NS) = 6/10
Therefore, the probability of not studying is 6/10.
P(Pa) = P(S,P) + P(NS,P)
= P(S) x P(P) + P(NS) x P(P)
= 4/10 x 2/3 + 6/10 x 1/5
= 8/30 + 6/50
= 29/75
~0.39
Therefore, the probability of passing the exam is 29/75 or approx 0.39.

b) If p = 0.75, find the probability that a student fails the examination.
P(S) = $\frac{?}{?}$ Write the fraction for 0.75.
P(NS) = 1 – P(S) Events S and NS are complementary.
= 1 – $\frac{?}{?}$
= $\frac{?}{?}$
P(F) = $\frac{?}{?}$ • $\frac{?}{?}$ + $\frac{?}{?}$ • $\frac{?}{?}$ Evaluate P(S, F) + P(NS, F).
= $\frac{?}{?}$
If the probability of studying is 0.75, then the probability of failing is $Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 2$ .
Answer:
P(NS) + P(S) = 1
P(NS) = 1 – P(S)
P(NS) = 1 – 3/4
P(NS) = 1/4
Therefore, the probability of not studying is 1/4.
P(Fa) = P(S,F) + P(NS,F)
= P(S) x P(F) + P(NS) x P(F)
= 3/4 x 1/3 + 1/4 x 4/5
= 3/12 + 4/20
= 9/20
= 0.45
Therefore, the probability of failing the exam is 9/20 or approx is 0.45

Math in Focus Course 3B Practice 11.4 Answer Key

State whether each event is a dependent or independent event.

Question 1.
Drawing 2 red balls randomly, one at a time without replacement, from a bag of six balls.
Answer: Dependent event

Question 2.
Tossing a coin twice.
Answer: Independent event

Question 3.
Reaching school late or on time for two consecutive days.
Answer: Independent event

Question 4.
Flooding of roads during rainy or sunny days.
Answer: Dependent event

Draw the tree diagram for each compound event.

Question 5.
2 balls are drawn at random, one at a time without replacement, from a bag of 3 green balls and 18 red balls.
Answer:
$Math in Focus Course 3B Practice 11.4 Answer Key_5$

Question 6.
The probability of rain on a particular day is 0.3. If it rains, then the probability that Renee goes shopping is 0.75. If it does not rain, then the probability that she goes jogging is 0.72. Assume that shopping and jogging are mutually exclusive and that rain and no rain are complementary.
Answer:
The probability of raining is 0.3, and the probability of raining P(R) and the probability of not raining P(NR) is equal to 1, then
P(NR) + P(R) = 1
P(NR) = 1 – P(R)
P(NR) = 1 – 0.3
P(NR) = 0.7
Therefore, the probability of not raining is 0.7
The probability of shopping when rain comes is 0.75, and the probability of shopping P(S), when it rains and the probability of jogging P(J) during rainy day is equal to 1, then
P(S) + P(J) = 1
P(J) = 1 – P(S)
P(J) = 1 – 0.75
P(J) = 0.25
Therefore, the probability of jogging when rains is 0.25
The probability of jogging when the rain did not come is 0.72, and the probability of shopping P(S) when it did not rain and the probability of jogging P(J) during a non-rainy day is equal to 1, then
P(S) + P(J) = 1
P(S) = 1 – P(J)
P(S) = 1 – 0.72
P(S) = 0.28
Therefore, the probability of shopping when it did not rain is 0.28.
$Math in Focus Course 3B Practice 11.4 Answer Key_6$

Solve. Show your work.

Question 7.
Geraldine has a box of 13 colored pens: 3 blue, 4 red, and the rest black. What is the probability of drawing two blue pens randomly, one at a time without replacement?
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 6$
Answer:
P(B, B) = P(B) x P(B)
= 3/13 x 2/12
= 1/26.
Therefore, the probability of picking blue pens twice is 1/26.

Question 8.
A box contains 8 dimes, 15 quarters, and 27 nickels. A student is to randomly pick two items, one at a time without replacement, from the bag. Find the probability that 2 quarters are picked.
Answer:
P(Q,Q) = P(Q) x P(Q)
= 15/50 x 14/49
= 210/2450
= 3/35
Therefore, the probability of picking a quarter twice is 3/35.

Question 9.
There are 9 green, 2 yellow, and 5 blue cards in a deck. Players A and B each randomly pick a card from the deck. Player A picks a card first before player B picks. Find the probability that both players pick the same color cards.
Answer:
P(B,B) or P(Y,Y) or P(G,G) = P(B,B) + P(Y,Y) + P(G,G)
= P(B) x P(B) + P(Y) x P(Y) + P(G) x P(G)
= 5/16 x 4/15 + 2/16 x 1/15 + 9/16 x 8/15
= 20/240 + 2/240 + 72/240
= 94/240
= 47/120
Therefore, the probability of obtaining a matching color of cards is 47/120.

Question 10.
The probability diagram below shows the probability of Xavier going to library or park depending if the weather is sunny or rainy. The probability of rain on a particular day is denoted by a. Assume that going to the library and going to the park are mutually exclusive and complementary.
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 7$
a) If a = 0.3, find the probability that Xavier goes to the park on any day.
Answer:
To convert decimals into fractions,
= 0.3/1
Multiplying both numerator and denominator to 100
= 0.3/1 x 100/100
= 30/100
= 3/10
Therefore, 0.3 is 3/10 in fractional form.
P(S) + P(R) = 1
P(S) = 1 – P(R)
P(S) = 1- 3/10
P(S) = 7/10
Therefore, the probability of having a sunny day is 7/10.
P(L) + P(P) = 1
P(P) = 1 – P(L)
P(P) = 1 – 1/5
P(P) = 4/5
Therefore, the probability of going to the park during the sunny day is 4/5.
P(S,P) or P(R,P) = P(S,P) + P(R,P)
= P(S) x P(P) + P(R) x P(P)
= 7/10 x 4/5 + 3/10 x 1/3
= 28/50 + 3/30
= 33/30
= 0.66
Therefore, the probability of going to park at any day is 0.66.

b) If a = 0.75, find the probability that he goes to the library on any day.
Answer:
Converting the decimals into fractions,
= 0.75/1
Multiply both the numerator and denominator to 100,
= 0.75/1 x 100/100
= 75/100
= 3/4
Therefore, 3/4 is the fractional form of 0.75
P(S) + P(R) = 1
P(S) = 1 – P(R)
P(S) = 1 – 3/4
P(S) = 1/4
Therefore, the probability of having a sunny day is 1/4.
P(S,L) or P(R,L) = P(S,L) + P(R,L)
= P(S) x P(L) + P(R) x P(L)
= 1/4 x 1/5 + 3/4 x 2/3
= 1/20 + 6/12
= 11/20
= 0.55
Therefore, the probability of going to library at any day is 0.55.

Question 11.
There are 15 apples in a fruit basket. 6 of them are red apples and the rest green apples. Two apples are picked randomly, one at a time without replacement.
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 8$
a) Draw a tree diagram to represent the possible outcomes.
Answer:
$Math in Focus Course 3B Practice 11.4 Answer Key_11a$

b) Find the probability of picking a green apple and then a red apple.
Answer:
P(G,R) = P(G) x P(R)
= 9/15 x 6/14
= 54/210
= 9/35
Therefore, the probability of picking a green apple then a red apple is 9/35.

c) Find the probability of picking two green apples.
Answer:
P(G,G) = P(G) x P(G)
= 9/15 x 8/14
= 72/210
= 12/35
Therefore, the probability of getting two green apples is 12/35.

d) Find the probability of picking two red apples.
Answer:
P(R,R) = P(R) x P(R)
= 6/15 x 5/14
= 30/210
= 1/70
Therefore, the probability of obtaining two red apples is 1/7.

Question 12.
There are 8 people in a room: 3 of them have red hair, 2 have blonde hair, and the rest have dark hair. Two people are randomly selected to leave the room, one after another, and they do not re-enter the room.
a) Draw a tree diagram to represent the possible outcomes.
Answer:
$Math in Focus Course 3B Practice 11.4 Answer Key_12a$

b) What is the probability of a person with dark hair leaving the room first?
Answer:
P(D,D) or P(D,R) or P(D,B) = P(D,D) + P(D,R) + P(D,B)
= P(D) x P(D) + P(D) x P(R) + P(D) x P(B)
= 3/8 x 2/7 + 3/8 x 3/7 + 3/8 x 2/7
= 6/56 + 9/56 + 6/56
= 21/56
= 3/8
Therefore, the probability that a dark haired person is the first one to leave is 3/8

c) What is the probability of a person with red hair leaving the room, followed by a person with blonde hair?
Answer:
P(R,B) = P(R) x P(B)
= 3/8 x 2/7
= 6/56
= 3/28
Therefore, the probability that a red haired person is the first one to leave then a blonde is 3/28.

d) What is the probability of two people with the same hair color leaving the room?
Answer:
P(D,D) or P(R,R) or P(B,B) = P(D,D) + P(R,R) + P(B,B)
= P(D) x P(D) + P(R) x P(R) + P(B) x P(B)
= 3/8 x 2/7 + 3/8 x 2/7 + 2/8 x 1/7
= 6/56 + 9/56 + 2/56
= 17/56
Therefore, the probability that a dark haired person is the first to leave is 17/56.

Question 13.
Along a stretch of road there are 2 traffic light intersections. Having red or green light for the first intersection is equally likely. Having a red light at the second intersection is twice as likely as a green light, if the first intersection traffic light was red. What is the probability of having a red light on the first intersection and a green light on the second intersection? Draw a tree diagram to show the possible outcomes.
Answer:
$Math in Focus Course 3B Practice 11.4 Answer Key_13a$
P(R,G) = P(R) x P(G)
= 1/2 x 1/3
= 1/6
Therefore, the probability of encountering a red then a green light is 1/6.

Question 14.
To get to work, Mr. Killiney needs to take a train and then a bus. The probability that the train breaks down is 0.1. When the train breaks down, there is a 0.7 probability that the bus will be overcrowded. When the train is operating normally, there is a 0.2 probability that the bus will be overcrowded. What is the probability of getting a seat in the bus? Draw a tree diagram to show the possible outcomes.
Answer:
P(SW) + P(W) = 1
P(W) = 1 – P(SW)
P(W) = 1 – 0.1
P(W) = 0.9
Therefore, the probability of the train working normally is 0.9.
P(NF) + P(F) = 1
P(NF) = 1 – P(F)
P(NF) = 1 – 0.7
P(NF) = 0.3
Therefore, the probability that the bus is not full is 0.3 when the train is not working properly.
P(NF) + P(F) = 1
P(NF) = 1- P(F)
P(NF) = 1 – 0.2
P(NF) = 0.8
Therefore, the probability that the bus is not full is 0.8 when the train is not working properly.
$Math in Focus Course 3B Practice 11.4 Answer Key_15$
P(SW,NF) or P(W,NF) = P(SW) x P(NF) + P(W) x P(NF)
= 0.1 x 0.3 + 0.9 x 0.8
= 0.03 + 0.72
= 0.75
Therefore, the probability that the bus is not full is 0.75.

Brain @ Work

Question 1.
If there are 12 green and 6 red apples, find the probability of randomly choosing three apples of the same color in a row, without replacement. Show your work.
Answer:
P(G,G,G) = P(G) x P(G) x P(G)
= 12/18 x 11/17 x 10/16
= 55/204
Therefore, the probability of obtaining green apples is 55/204.
P(R,R,R) = P(R) x P(R) x P(R)
= 6/18 x 5/17 x 4/16
= 5/204
Therefore, the probability of obtaining red apples is 5/204.
P(R,R,R) or P(G,G,G) = P(R) x P(R) x P(R) + P(G) x P(G) x P(G)
= 5/204 + 55/204
= 60/204
= 5/17
Therefore, the probability of obtaining of picking 3 apples of a matching color is 5/17.

Question 2.
William has five $1 bills, ten $10 bills, and three $20 bills in his wallet. He picks three bills randomly in a row, without replacement. What is the probability of him picking three of the same type of bills? $how your work.
Answer:
P($20,$20,$20) = P($20) x P($20) x P($20)
= 3/18 x 2/17 x 1/16
= 6/4896
= 1/816
Therefore, the probability of obtaining three $20 bills is $1/816.
P($10,$10,$10) = P($10) x P($10) x P($10)
= 10/18 x 9/17 x 8/16
= 720/4896
= 5/34
Therefore, the probability of obtaining three $10 bills is $5/34.
P($1,$1,$1) = P($1) x P($1) x P($1)
= 5/18 x 4/17 x 3/16
= 60/4896
= 5/408
Therefore, the probability of obtaining three $1 bills is 5/408.
P($20,$20,$20) or P($10,$10,$10) or P($1,$1,$1) = P($20,$20,$20) + P($10,$10,$10) + P($1,$1,$1)
= 1/816 + 5/34 + 5/408
= 131/816
Therefore, the probability of having three bills of a matching type is 131/816.

Question 3.
Daniel plans to visit Australia. Whether he goes alone or with a companion is equally likely. If he travels with a companion there is a 40% chance of joining a guided tour. If he travels alone, there is an 80% chance of joining a guided tour.
$Math in Focus Grade 8 Chapter 11 Lesson 11.4 Answer Key Dependent Events 9$
a) What is the probability of traveling with a companion and not joining a guided tour?
Answer:
P(NT) + P(T) = 1
P(NT) = 1 – P(T)
P(NT) = 1 – 0.4
P(NT) = 0.6
Therefore, the probability of not going to tour is 0.6 if he travels with others.
P(NT) + P(T) = 1
P(NT) = 1 – P(T)
P(NT) = 1 – 0.8
P(NT) = 0.2
Therefore, the probability of not going to tour is 0.2 if he travels without the other.
P(C,NT) = P(C) x P(NT)
= 0.5 x 0.6
= 0.3
Therefore, the probability of travelling with others and not going with a tour is 0.3.

b) What is the chance of joining a guided tour?
Answer:
P(C,T) or P(WC,T) = P(C) x P(T) + P(WC) x P(T)
= 0.5 x 0.4 + 0.5 x 0.8
= 0.2 + 0.4
= 0.6
Therefore, the probability of being in a tour is 0.6