Which value is equivalent to Check the picture

Of the 50 U.S. states, 4have names that start with the letter W . What percentage of U.S. states have names that start with the

Verizon [17]

8% of US states start with W

4 0

2 years ago

Write the product (4 + i)(4 − i) in the form a + bi. A. 16 − i B. 16 − i2 C. 17 D. 8

bonufazy [111]

Answer:

A.16–i is correct.

Step-by-step explanation:

(4+i)(4–i). [(a+b)(a–b)=(a²–b²)]

(4²–i²)=16–i

I hope this answer is correct so please vote my answer

3 0

3 years ago

For each equation, determine whether it represents a direct variation, an inverse variation, or neither. Find the constant of va

rewona [7]

Both inverse variation’s. K=constant=9,6

6 0

3 years ago

Which statement describes the system of equations

Contact [7]

Answer:

2y-2x-2y= - 22x = 0x + 2y = 1

Step-by-step explanation:

7 0

3 years ago

One study reports that 34% of newly hired MBAs are confronted with unethical business practices during their first year of empl

MaRussiya [10]

Answer:

$z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364$

$p_v =2*P(Z$

The p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ 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 proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.

Step-by-step explanation:

1) Data given and notation

n=116 represent the random sample taken

X represent the number graduates from the previous year claim to have encountered unethical business practices in the workplace

$\hat p=0.28$ estimated proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace

$p_o=0.34$ is the value that we want to test

$\alpha=0.05$ represent the significance level

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 is 0.34 or no.:

Null hypothesis: $p=0.34$

Alternative hypothesis: $p \neq 0.34$

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:

$z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364$

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 next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z$

The p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ 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 proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.

5 0

3 years ago