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oksano4ka [1.4K]

3 years ago

6

I dont understand how to write this 2 problems out in equivalent form

1 answer:

deff fn [24]3 years ago

8 0

Just do 1/8 times 1/8. Hope this helped !!

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Please help me with these problems.

kkurt [141]

The first question is A and number 2 is B

7 0

4 years ago

mark purchased a new motorcycle for $23,000. F the value of the motorcycle decreases by 151 each year, what will it be worth in

HACTEHA [7]

151 * 5 = 755
23,000 - 755 = 22,245

It'll be worth $22,245 in 5 years.

4 0

4 years ago

Which table shows a nonlinear relationship? A. x|5|4|0|. Y|15|11|5|. B. X|-1|0|1|. Y|15|18|21|. C. X|5|7|9|. Y|15|11|5|. D. X|5

Damm [24]

Answer: c

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A school held a jump-roping contest. Diego jumped rope for 20 minutes.

snow_tiger [21]

Answer:

Jada 15

Lin 24

Noah 9

Diego 20

Step-by-step explanation:

Hope I helped

4 0

3 years ago

The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T

Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample. In this problem, $\sigma = 0.25$

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.1 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.1\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.1}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}$

$n = 24.01$

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.06 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.06\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.06}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}$

$n = 66.7$

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.05 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.05\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.05}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}$

$n = 96.04$

Rounding up, 97

8 0

3 years ago