**Answer:**

Check the explanation below.

**Step-by-step explanation:**

Hello!

The study variable is X: one-time online fundraising that a nonprofit organization has received.

Sample:

n= 50

sample mean X[bar]= $62

Standard deviation S= $9

You are interested in testing the average one-time donation is different than $60. Symbolically: μ ≠ 60

The hypothesis is:

H₀: μ = 60

H₁: μ ≠ 60

You didn't stat any significance level, I'll use a level of 5%

The critical region and the p-value for this test are two-tailed. Since all the available information is from the sample, assuming that the variable has normal distribution I've chosen the Student-t statistic:

t= <u> X[bar] - μ </u>~ t

S/√n

t= <u> 62 - 60 </u>= 1.57

9/√50

The p-value for t=1.57 is 0.1228

Since the p-value: 0.1228 is greater than the significance level of 5%, the decision is to reject the null hypothesis. This means that the average one-time gift online fundraising that a nonprofit organization received is equal to $60.

I hope it helps!

**Answer:**

This question is incomplete.

**Step-by-step explanation:**

Where are the list of equivalent values?

One list might be 98+1414+918 (because there are 7 14's either side of the 1414.)

**Answer:**

its decreasing. .........

**Answer:**

4:7

**Step-by-step explanation:**

3:5=6:10/2=3:5

9:15=9:15/3=3:5

**Answer:**

**Step-by-step explanation:**

from the picture:

QP = QR

and

QR = RS

so

PQ + RS = QS