What's 15/.75 and 99/162? Show works please. :)

Match each problem to its solution 8,8.2,9,56.25

Mekhanik [1.2K]

The computation shows that the amount will be:

$56.25.
$9
8
8.2 pounds

<h3>How to compute the values?</h3>

Jonah earns $12.50 per hour

He works for 4.5 hours

Amount he earned will be:

= 12.50 × 4.5 = $55.25

Items worth is $41.23

Amount paid is $50.23

Extra amount paid will be:

= 50.23 - 41.23

= $9

Length of cloth available is 20 m

Length of one long-sleeved blouse is 2.5m

Number of blouses that can be made will be:

= 20/2.5 = 8

Potatoes weight is 3.2 pounds

Oranges weight is 2.3 pounds

Onions weight is 2.7 pounds

Total weight of the purchase will be:

= 3.2 + 2.3 + 2.7 = 8.2 pounds

Learn more about the computation of amount on:

brainly.com/question/20038451

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<u>Complete question:</u>

Match each problem to its solution. 8 8.2 9 56.25 Jonah earns $12.50 per hour as a marketing intern. If he works for 4.5 hours, how many dollars does he earn? arrowRight Jessica purchased items worth $41.23 at the local grocery store. The cashier made a mistake while ringing up her items, and Jessica ended up paying $50.23. How many extra dollars did she pay? arrowRight Josie has 20 meters of cloth. She needs to make long-sleeved blouses with that cloth. If one long-sleeved blouse requires 2.5 meters of cloth, how many blouses can she make with the material? arrowRight Jeremy bought 3.2 pounds of potatoes, 2.3 pounds of oranges, and 2.7 pounds of onions. How many pounds does his purchase weigh? arrowRight

3 0

1 year ago

Nitrates are groundwater contaminants derived from fertilizer, septic tank seepage and other sewage. Nitrate poisoning is partic

o-na [289]

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

b) $n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

Part a

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Since we don't have prior information about the population proportion we can use ase estimator $\hat p =0.5$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

Part b

For thi case the estimator for p is $\hat p =0.07$

$n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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2 years ago

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