5. What is the value of x in the triangle at the right? A. 23 B. 32 C. 125 D. 148

6(-2.3x-5)+(4x+11) What is the answer? Thanks!

cricket20 [7]

That’s what I got for the answer

8 0

3 years ago

A recent article reported that a job awaits only one in three new college graduates. The major reasons given were an overabundan

nikitadnepr [17]

Answer:

$z=\frac{0.4 -0.33}{\sqrt{\frac{0.33(1-0.33)}{200}}}=2.105$

$p_v =P(z>2.105)=0.018$

If we compare the p value obtained and the significance level given $\alpha=0.01$ 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 1% of significance the proportion of students that had job at the school mentioned is not significantly higher then 0.33 .

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X=80 represent the number of students that had jobs

$\hat p=\frac{80}{200}=0.4$ estimated proportion of students that had jobs

$p_o=0.3333$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of students that have jobs at the school mentioned is higher than the reported value at the article:

Null hypothesis: $p\leq 0.33$

Alternative hypothesis: $p > 0.33$

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:

$z=\frac{0.4 -0.33}{\sqrt{\frac{0.33(1-0.33)}{200}}}=2.105$

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

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

$p_v =P(z>2.105)=0.018$

If we compare the p value obtained and the significance level given $\alpha=0.01$ 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 1% of significance the proportion of students that had job at the school mentioned is not significantly higher then 0.33 .

7 0

3 years ago

Find the value of the expression below. Express your answer in scientific notation.

nirvana33 [79]

The value of the expression is $8.8 \times 10^{-2}$ .