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klasskru [66]
4 years ago
9

The amounts of paper waste generated in a region during two years were 41,300,000 tons and 50,500,000 tons. What was the total p

aper waste generated over these two years in the region, expressed in scientific notation? (1 point) 91.8 ⋅ 107 tons 91.8 ⋅ 1014 tons 9.18 ⋅ 1014 tons 9.18 ⋅ 107 tons
Mathematics
2 answers:
Viktor [21]4 years ago
6 0

It's easier to add these first and then change to SN.

41,300,000 + 50,500,000 = 91,800,000

SN needs to change to this format:

9.18 x 10^7

saul85 [17]4 years ago
5 0

Answer: 9.18\times10^{7}\text{ tons}

Step-by-step explanation:

Given : The amounts of paper waste generated in a region during two years were 41,300,000 tons and 50,500,000 tons.

Then, the total paper waste generated over these two years in the region :-

41,300,000+50,500,000=91,800,000

In scientific notation,

91,800,000=9.18\times10^{7}

Hence, the total paper waste generated over these two years in the region =

9.18\times10^{7}\text{ tons}

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Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

n = 30 -- sample size

\bar x = 50 -- sample mean

\sigma = 10 --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

\sigma_x = \frac{\sigma}{\sqrt n}

\sigma_x = \frac{10}{\sqrt {30}}

\sigma_x = \frac{10}{5.478}

\sigma_x = 1.825

The 95% confidence interval for the z value is:

z = 1.960

Calculate margin of error (E)

E = z * \sigma_x

E = 1.960 * 1.825

E = 3.577

The confidence bound is:

Lower = \bar x - E

Lower = 50 - 3.577

Lower = 46.423

Lower = 46.4 --- approximated

Upper = \bar x + E

Upper = 50 + 3.577

Upper = 53.577

Upper = 53.6 --- approximated

<em>So, the 95% confidence interval is (46.4, 53.6)</em>

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

\sigma_x = \frac{\sigma}{\sqrt n}

\sigma_x = \frac{10}{\sqrt {90}}

\sigma_x = \frac{10}{9.49}

\sigma_x = 1.054

The 95% confidence interval for the z value is:

z = 1.960

Calculate margin of error (E)

E = z * \sigma_x

E = 1.960 * 1.054

E = 2.06584

The confidence bound is:

Lower = \bar x - E

Lower = 50 - 2.06584

Lower = 47.93416

Lower = 47.9 --- approximated

Upper = \bar x + E

Upper = 50 + 2.06584

Upper = 52.06584

Upper = 52.1 --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

n = 30     E = 3.577

In (b), we have:

n = 90    E = 2.06584

<em>Notice that the margin of error decreases when the sample size increases.</em>

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