Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 2 Review Test to finish your assignments.

## Math in Focus Grade 8 Course 3 A Chapter 2 Review Test Answer Key

**Concepts and Skills**

**Tell whether each number is written correctly in scientific notation. If incorrectly written, state the reason.**

Question 1.

10 • 10^{2}

Answer:

10 • 10^{2 }is correct in scientific notation, because the given notation is equal to 10.

Question 2.

0.99 • 10^{12}

Answer:

The 0.99 • 10^{12 }is incorrect in scientific notation.

The coefficient is less than 1

Question 3.

1.4 • 10^{2}

Answer:

1.4 • 10^{2 }is Correct in scientific notation, because the given notation is in between 1 to 10.

Question 4.

0.4 • 10^{25}

Answer:

0.4 • 10^{25 }is incorrect in scientific notation.

Because the coefficient is less than 1.

**Write each number in scientific notation.**

Question 5.

714,000

Answer:

714,000 in the scientific notation is

7.14 •10⁵

Question 6.

0.00087

Answer:

0.00087 in the scientific notation is 0.87 • 10^{-3}

**Write each number in standard form.**

Question 7.

3.46 • 10^{2}

Answer:

3.46 • 10²

10² = 100

3.46 × 100 = 346.

Question 8.

5.4 • 10^{4}

Answer:

5.4 • 10⁴

5.4 × 10,000

= 54,000

**Identify the greater number in each pair of numbers.**

Question 9.

7.8 • 10^{-5} and 5.4 • 10^{-7}

Answer:

7.8 • 10^{-5} is greater than the 5.4 • 10^{-7}

Because 7.8 • 10^{-5} has a big exponent.

Question 10.

1.4 • 10^{-5} and 6 • 10^{-4}

Answer:

6 • 10^{-4} is greater than the 1.4 • 10^{-5}

Because 6 • 10^{-4} has a big exponent.

Question 11.

6.5 • 10^{-15} and 9.3 • 10^{-12}

Answer:

9.3 • 10^{-12} is greater than the 6.5 • 10^{-15}

Because 9.3 • 10^{-12} has a big exponent.

Question 12.

3.5 • 10^{-2} and 4 • 10^{-3}

Answer:

3.5 • 10^{-2} is greater than the 4 • 10^{-3}

Because 3.5 • 10^{-2} has a big exponent.

**Evaluate. Write your answer in scientific notation. Round the coefficient to the nearest tenth.**

Question 13.

2.44 • 10^{3} + 1.9 • 10^{5}

Answer:

2.44 • 10^{3} + 1.9 • 105

Rewrite the numbers to have the same power of 10

0.0244 • 10⁵ + 1.9 • 10⁵

Now add the numbers

0.0244 + 1.9 = 1.9244

Rewrite the number in scientific notation is

1.9244 • 10⁵

1.9244 rounded to the nearest tenth is 1.9

Question 14.

3.12 • 10^{-3} – 3 • 10^{-3}

Answer:

3.12 • 10^{-3} – 3 • 10^{-3}

Now subtract the numbers

(3.12 – 3) • 10^{-3}

Rewrite the number in scientific notation is

0.12 • 10^{-3}

Question 15.

2.4 • 10^{-2} • 5 • 10^{-1}

Answer:

2.4 • 10^{-2} • 5 • 10^{-1}

Here bases are equal powers should be added.

2.4 • 5 • 10^{-3}

12 • 10^{-3}

Question 16.

3.2 • 10^{8} ÷ (1.6 • 10^{4})

Answer:

3.2 • 10^{8} ÷ (1.6 • 104)

3.2/1.6 × 10⁸/10⁴

3.2/1.6 × 10⁴

2 • 10⁴

**Express each of the following in prefix form. Choose the most appropriate unit.**

Question 17.

2.8 • 10^{3} meters

Answer:

2.8 • 10^{3} meters

We have to using the prefix form

2.8 • 10¹ • 10²

2.8 • 10¹ centimeters

28 centimeters

Question 18.

1.5 • 10^{-6} meter

Answer:

1.5 • 10^{-6} meter

We have to use the prefix form

1.5 • 10^{-6} meter

1.5 micrometers

Question 19.

6.4 • 10^{9} bytes

Answer:

6.4 • 10^{9} bytes

We have to use the prefix form

64 • 10^{8} bytes

Question 20.

4.8 • 10^{-9} gram

Answer:

4.8 • 10^{-9} gram

We have to use the prefix form

4.8 • 10^{-6} •10^{-3} gram

4.8 •10^{-3} micrograms

0.0048 micrograms

**Problem Solving**

**The table shows the length of two organisms. Use the table to answer questions 21 to 23.**

Question 21.

Which organism is longer?

Answer:

Given that the length of eriophyid mite = 250 micrometers

The length of patiriella parvivipara = 5 millimeter.

1 micrometer = 0.001 millimeters

250 micrometers = 0.001 • 25 = 0.025 millimetres

Therefore, The length of patiriella parvivipara is longer.

Question 22.

Express the length of the eriophyid mite in millimeters.

Answer:

Given that the length of the eriophyid mite 250 micrometers

1 micrometer = 0.001 millimeters

250 micrometers = 0.001 • 25 = 0.025 millimeters

Therefore the length of the eriophyid mite in millimeters is 0.025

Question 23.

Write each length in scientific notation using the basic unit.

Answer:

Given that the length of eriophyid mite = 250 micrometers

0.025 • 10^{4}

The length of patiriella parvivipara = 5 millimeter.

5 millimetres = 0.5 • 10¹

**The top five materials used in the automotive industry in the United States in 2000 are as shown in the table. Use the table to answer questions 24 to 26. Write your answers in scientific notation. Round coefficients to the nearest tenth.**

Question 24.

How much more plastic was used than aluminum?

Answer:

Given that the plastic material used in the automotive industry is 46240 = 46.240 • 10³

The aluminum Material used in the automotive industry is 11,320 = 11.320 • 10³

Plastic material – aluminium material = 46.240 • 10³ – 11.320 • 10³ = 34.92 = 3.492 • 10¹

Plastic material is 3.492 • 10¹ consumption is used more than aluminum.

34920 rounded to the nearest tenth is 34920

Question 25.

How much more steel was used than glass?

Answer:

Given that the steel material used in the automotive industry is 9.894 • 10⁷

9.894 • 10⁷

The glass material used in the automotive industry is 54,17,000 = 0.5417000 • 10⁷

Steel material – glass material = 9.894 • 10⁷ – 0.5417000 • 10⁷ = 9.3523 = 93.523 • 10^{-1}

93.523 • 10^{-1} rounded to the nearest tenth is 93.5 • 10^{-1}

Question 26.

Find the total consumption of these materials used by the automotive industry in 2000.

Answer:

Given that the plastic material used in the automotive industry is 46,240 = 46.240 • 10³

The aluminum material used in the automotive industry is 11,320 = 11.320 • 10³

The steel material used in the automotive industry is 9.894 • 10⁷

The glass material used in the automotive industry is 54,17,000 = 0.5417000 • 10⁷

The rubber material used in the automotive industry is 2.86 • 10⁶

Total consumption = 46.240 • 10³ + 11.320 • 10³ + 9.894 • 10⁷ + 0.5417000 • 10⁷ + 2.86 • 10⁶ = 70.8547 • 10²⁶

70.8547 • 10²⁶ rounded to the nearest tenth is 70.85 • 10²⁶

Question 27.

The table shows the weights of some animals. Round your answers to the nearest tenth.

a) About how many times greater than the weight of a walrus is the weight of a hippopotamus?

Answer:

Given that the weight of a walrus = 4.4 • 10³

And the weight of a hippopotamus = 9.9 • 10³

weight of hippopotamus – weight of walrus = 9.9 • 10³ – 4.4 • 10³ = 5.5

Weight of a hippopotamus is 5.5 times greater than that of a walrus.

5.5 rounded to the nearest tenth is 5.5

b) About how many times greater than the weight of a hippopotamus is the weight of an African bush elephant?

Answer:

Given that the weight of a hippopotamus = 9.9 • 10³

The weight of a African bush elephant = 2.706 • 10⁴ = 27.06 • 10³

Weight of African bush elephant – weight of a hippopotamus = 27.06;• 10³ – 9.9 • 10³ = 17.16

The weight of an African bush elephant is 17.16 times greater than the hippopotamus.

17.16 rounded to the nearest tenth is 17.2

Question 28.

The floor of the Palace of the Parliament in Romania measures 8.9 • 10^{2} feet by 7.9• 10^{2} feet. The building reaches 2.82 • 10^{2} feet above ground, and 3.02 • 10^{2} feet below ground. Use the formula for the volume of a rectangular prism to estimate the volume inside the palace.

Answer:

Length: 8.9 • 10^{2} feet We are given the dimensions of the rectangular prism:

Width: 7.9 • 10^{2} feet

Above height: 2.82 • 10^{2} feet

Below height: 3.02 • 10^{2} feet

Length • Width • (Above height + Below height) We determine the volume of the prism:

= 8.9 • 10^{2} • 7.9 • 10^{2} • (2.82 • 10^{2} + 3.02 • 10^{2})

= 8.9 • 10^{2} 7.9 • 10^{2} • [(2.82 + 3.02) • 10^{2}]

= 8.9 • 10^{2} • 7.9 • 10^{2} • 5.84 • 10^{2}

= 8.9 • 7.9 • 5.84 • 10^{2} • 10^{2} • 10^{2}

= 410.6104 • 10^{2+2+2}

= 410.6104 • 10^{6}

= 4.106104 • 10^{8}

≈ 4.1 • 10^{8} cube feet

Question 29.

The Stockholm Globe Arena in Sweden resembles a hemisphere. It has an inner diameter of about 1.1 • 10^{2} meters. Find the approximate volume of the building. Use 3.14 as an approximation for π.

Answer:

Given that the inner diameter of a hemisphere is 1.1 • 10².

Assume that the building is in the shape of a hemisphere.

D = 2r

1.1 • 10² = 2r

r = 1.1 • 10²/2 = 110/2 = 55

Volume of a hemisphere = ⅔ πr³ cubic units

= ⅔ • 3.14 • (55)³

= 348,278.33