Please answer asap 25 points

Susan and terry run a day care center Since they use susans house it was agreed that her share is twice as Terrys if they earn 2

Nadya [2.5K]

The easiest way to do this is to divide 225 by 3, so 225 / 3, and what you get as the sum, in this case it's 75, you would multiply the sum by 2, which equals 150, and that, would be Susan's share. So... 225 / 3 = 75. 75 * 2 = 150. $150 is the amount Susan gets. Hope this helps!

6 0

2 years ago

Х = 4 3х — у = -1 How to solve ?

elena-s [515]

3+ - 1 I hole this helped

7 0

2 years ago

A rectangular box is 5 in. wide, 5 in. tall, and 4 in. long. What is the diameter of the smallest circular opening through whi

Katyanochek1 [597]

Answer:

7 (7.07106781187)

Step-by-step explanation:

All you have to do is find the diameter of the rectangle from two farthest corners. To do this, use the Pythagorean Theorem

(5^2) + (5^2) = c^2

25 + 25 = c^2

50 = c^2

7 ≈ c

8 0

3 years ago

According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bother

k0ka [10]

Answer:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

$p_v =P(z>7.124)=5.24x10^{-13}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

$\hat p=\frac{331}{502}=0.659$ estimated proportion of people who would not be bothered if the NAtional security agency

$p_o=0.5$ 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

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than the majority of 0.5:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

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:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

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 right tailed test the p value would be:

$p_v =P(z>7.124)=5.24x10^{-13}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.

4 0

3 years ago

What is the approximate estimation for 29.942 + 13.433

Nana76 [90]

Answer:

the answer would be it is approximately 43.85

(did addition)

6 0

2 years ago

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