PLEASE HELP THIS IS THE LAST QUESTION!!

This makes no sense, help pls

stiv31 [10]

Please mark me branliest!

**Answer**

16 if you want exact, 20 if you want a little over.

**Explanation**

So check this... it's $50 per student that attends the camp, right?

(Nod if you're with me.)

The dude buys 5 dozen donuts for 16 days. 50 x 16 = 800

5 dozen is the same as 50 donuts.

You multiply to get the total amount of donuts he buys in the total days (16).

Each 10 donuts costs 10 dollars.

If the dude buys 800 donuts it cost him $800.

And it costs 50 dollars for someone to go to camp.

So you divide 800/50 = 16 students

Now the problem says "In order to break even" I'm not sure what that means. But I'm assuming it doesn't want 16 then, because the price of 16 students would be the same (even) as the price of donuts that Mr. Brook buys. In that case it would be 20.

You are very welcome!!

8 0

3 years ago

What is this answer?

Shtirlitz [24]

Answer:

1/2

Step-by-step explanation:

your mom :|

4 0

2 years ago

Read 2 more answers

A bookstore has 5 boxes of 10 in each box on Monday the bookstore sold 20 books how many books remain to be sold?

AlekseyPX

Answer: 30

Step-by-step explanation: five boxes with ten each which means there's a total of 50 item if they sold 20 you would do 50 -20=30

8 0

2 years ago

The standard IQ test has a mean of 97 and a standard deviation of 17. We want to be

densk [106]

Let's call our estimate x. It will be the average of n IQ scores. Our average won't usually exactly equal the mean 97. But if we repeated averages over different sets of tests, the mean of our estimate the average would be the same as the mean of a single test,

μ = 97

Variances add, so the standard deviations add in quadrature, like the Pythagorean Theorem in n dimensions. This means the standard deviation of the average x is

σ = 17/√n

We want to be 95% certain

97 - 5 ≤ x ≤ 97 + 5

By the 68-95-99.7 rule, 95% certain means within two standard deviations. That means we're 95% sure that

μ - 2σ ≤ x ≤ μ + 2σ

Comparing to what we want, that's means we have to solve

2σ = 5

2 (17/√n) = 5

√n = 2 (17/5)

n = (34/5)² = 46.24

We better round up.

Answer: We need a sample size of 47 to be 95% certain of being within 5 points of the mean

6 0

2 years ago

A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school

nekit [7.7K]

Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let $p_1$ = proportion of Illinois high school freshmen who have used anabolic steroids.

$p_2$ = proportion of Illinois high school seniors who have used anabolic steroids.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of high school freshmen who have used anabolic steroids = $\frac{34}{1679}$ = 0.0203

$\hat p_2$ = sample proportion of high school seniors who have used anabolic steroids = $\frac{24}{1366}$ = 0.0176

$n_1$ = sample of high school freshmen = 1679

$n_2$ = sample of high school seniors = 1366

So, the test statistics = $\frac{(0.0203-0.0176)-(0)}{\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366} } }$

= 0.545

The value of z test statistics is 0.545.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

3 0

2 years ago