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3 years ago

Find the value of c.

Mathematics

1 answer:

hodyreva [135]3 years ago

7 0

Yeah, it is 16, the way they worked it out is correct also

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Find the value of RT

Elis [28]

Answer:

$RT=7$

Step-by-step explanation:

$RQ*RP=RS*RT$

$9*(9+13)=11(11+x)$

$9*\frac{22^2}{11} =\frac{11}{11} =(11+x)$

$18=11+x$ ⇒ $7$

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 hope it helps..

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3 0

3 years ago

HELP I AM TIMED!!!!!!!!!!!!

ExtremeBDS [4]

Answer:

It is the last one 6a-4b+11

Explanation:

Hope this helps if u need clarification plz ask:)

5 0

2 years ago

If you have $2500 how many board feet can you buy if the cost per board foot is $3.217

igor_vitrenko [27]

777 when rounded, or 777.1215418091389

3 0

3 years ago

Solve –15 = 4m – 7. –32 2 –12 –2

Shkiper50 [21]

Answer:

-2 =m

Step-by-step explanation:

–15 = 4m – 7

Add 7 to each side

–15+7 = 4m – 7+7

-8 = 4m

Divide each side by 4

-8/4 = 4m/4

-2 =m

4 0

3 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .

We have to find M.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 2.81\sqrt{\frac{0.12*0.88}{5615}}$

$M = 0.012$

7 0

3 years ago