Solve for x<br> A) -2<br> B) 2<br> C) -4<br> D) 4

dezoksy [38]

I think it would be (8)(-1) = -8. by adding the two equations you get 2x = 16 so x = 8 then y = 7 - x = -1. Therefore xy = -8.

Hope that helps

6 0

3 years ago

A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r

weqwewe [10]

Answer:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

$p_v =P(z$

If we compare the p value obtained and 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 americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

Step-by-step explanation:

1) Data given and notation

n=180 represent the random sample taken

X=125 represent the number of americans between 17 to 24 that not qualify for the military

$\hat p=\frac{125}{180}=0.694$ estimated proportion of americans between 17 to 24 that not qualify for the military

$p_o=0.75$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :

Null hypothesis: $p\geq 0.75$

Alternative hypothesis: $p < 0.75$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

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

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

$p_v =P(z$

If we compare the p value obtained and 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 americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

6 0

3 years ago

In a survey of 2,000 people who owned a certain type of car, 900 said they would buy that type of car again. What percent of th

soldi70 [24.7K]

Answer:

45%

Step-by-step explanation:

900/2,000 = 45

4 0

3 years ago

Six less than a number is twenty

nignag [31]