40 points PLZ HELP!

What is 8% of 37...........

docker41 [41]

Answer:

2.96

Explanation:

37/x=100/8

(37/x)*x=(100/8)*x - we multiply both sides of the equation by x

37=12.5*x - we divide both sides of the equation by (12.5) to get x

37/12.5=x

2.96=x

x=2.96

7 0

3 years ago

Read 2 more answers

If kendal payed $2,500 down and $150 each month for dealer one and at another dealer paid $3,000 down and $125 each month after

maxonik [38]

After 20 months. after 20 months the first dealer will cost 5500 and the second dealer will cost 5500
dealer one pays 3000 in monthly fines and 2500 down and dealer 2 pays 2500 in fines and 300 down

6 0

3 years ago

How I write tens thousands

Inga [223]

10,000.
or

0.0001 one ten-thousandth

6 0

3 years ago

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

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".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$