Which two points on the number line are 3 units away from 7?

What is this answer to this question

Mashutka [201]

Answer:

it would be 110. you add 80+30

6 0

3 years ago

Read 2 more answers

At a bake sale, students sold 3 muffins for every 8 cookies.the students sold 24 muffins altogether. how many cookies did they s

dsp73

64 Muffins

3:8
24:a

24/3 = y
y=8

a = 8 x y
a = 8 x 8
a = 64 muffins

3 0

3 years ago

Which of the following are solution to the equation below Check all that apply. <br> (4x-1)^2=11

Juliette [100K]

Answer:

I think it's E and F. Let me know if I am wrong.

Step-by-step explanation:

4 0

3 years ago

(PLS ANWSER FAST) Kyle has a storage box in the shape of a rectangular prism. the box has a square base with sides that are 3 ft

Dennis_Churaev [7]

V = length x width x height

In this problem:

length = 3

width = 3

height = 6

V = 3 x 3 x 6

V = 54 ft^3

Hope this helps!! :)

7 0

3 years ago

In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders o

melomori [17]

Answer:

A. We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B. Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

C. $z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

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

E. Fail to the reject the null hypothesis

F. So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

Step-by-step explanation:

Data given and notation

n=362 represent the random sample taken

X=33 represent the number of orders not accurate

$\hat p=\frac{33}{363}=0.0912$ estimated proportion of orders not accurate

$p_o=0.10$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

A: Write the claim as a mathematical statement involving the population proportion p

We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B: State the null (H0) and alternative (H1) hypotheses

Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

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 $p_o$ .

C: Find the test statistic

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

$z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D: Find the critical value(s)

Since is a bilateral test we have two critical values. We need to look on the normal standard distribution a quantile that accumulates 0.025 of the area on each tail. And for this case we have:

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

P value

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

E: Would you Reject or Fail to Reject the null (H0) hypothesis.

Fail to the reject the null hypothesis

F: Write the conclusion of the test.

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

6 0

3 years ago