Peter expected to earn $425 in a week. In fact, he earned $500. What was the percentage error? (Round to nearest percent)

25 is 36% of what number

ch4aika [34]

Answer:

x = 69.44

Step-by-step explanation:

We have, 36% × x = 25

or, 36/100 × x = 25

Multiplying both sides by 100 and dividing both sides by 36,

we have x = 25 × 100/36

x = 69.44

If you are using a calculator, simply enter 25×100÷36, which will give you the answer.

4 0

1 year ago

Marc buys three magazines for $7.50 at a school fundraiser. If all magizines in the fundraiser cost the same, what is the consta

mafiozo [28]

Answer:

90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you

Step-by-step explanation:

4 0

3 years ago

(4y-4)=(10x+65) answer please?

icang [17]

Answer:

Step-by-step explanation:

4y-4=10x+65, this can be expressed in many ways...

4y-10x-4-65=0

4y-10x-69=0 or

4y=10x+69

y=(10x+69)/4

3 0

2 years ago

HELP HELP HELP HELP!!I WILL MARK BRAINLIEST IF CORRECT! Find the tangents of the acute angles in the right triangle. Write each

Softa [21]

The answer for j box is 22 and for k boxs 15

4 0

3 years ago

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A ran

mart [117]

Answer:

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

$X=48$ number of vehicles classified as sports utility.

$n=500$ random sample taken

$\hat p=\frac{48}{500}=0.096$ estimated proportion of vehicles classified as sports utility vehicles.

$p$ true population proportion of vehicles classified as sports utility vehicles.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

The confidence interval would be given by this formula :

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070$

$0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121$

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

7 0

3 years ago