Please Help! Will Give Brainlest!

The yearbook staff receive 75 submissions for yearbook articles. Paul will accept 2/5 of all submissions. Currently, Paul plans

xz_007 [3.2K]

Answer 813.8

Step-by-step explanation:

Well 813.8 is the answer. Why? Lets talk it out if there are 75 submissions and it says Paul will accept 2/5 of the submissions so we have to do is 75- 2/5 will be 74.6 and multiply by 13 because 13 pages is in a article so we got 969.8 and then minus 156 because it asking us how many "more" pages they need so 813.8 is the answer if you subtract 969.8 and 156.

5 0

2 years ago

What is -2 1/2 x -3 1/3

Blizzard [7]

Answer:

6 1/6

Step-by-step explanation:

-2 1/2 * 3 1/3

-2 * -3 = 6

1/2 * 1/3 = 1/6

6 0

2 years ago

Please help with the questions in the image

allochka39001 [22]

Answer:

fghjk

Step-by-step explanation:

6 0

2 years ago

Find the circumference in feet.

Rama09 [41]

Answer:

C≈25.13ft

Step-by-step explanation:

Using the formulas

C=2πr

d=2r

Solving for C

C=πd=π·8≈25.13274ft

3 0

2 years ago

Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in

Alborosie

Answer:

a) Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

b) $z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

$p_v =P(Z$

c) So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=0.46$ estimated proportion of American families owning stocks or stock funds

$p_o=0.53$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:

Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

Part b

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

Part c

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

7 0

2 years ago