Please help

Emerson has an associate degree. Based on the bar chart below, how will his employment opportunities change from 2008-2018?

ANEK [815]

Answer:

19% change

Step-by-step explanation:

Since, Emerson has an associate degree,

His employment is 19% changed in 10 years, i.e. from 2008 to 2018.

Because the bar graph shown the percentage of change in employment from 2008 to 2018.

Also, Employment of associate degree is most affected in 10 years.

And those who have job training and work experience is least affected in these 10 years.

5 0

3 years ago

Divide 1-k^3 by k-1

UNO [17]

Answer:

k 2 + k + 1

Step By Step:

Cancel the common factor.

Then.

k 2 + k +1 by 1

Gets you k2+k+1

7 0

3 years ago

Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

Ket [755]

Answer:

Step-by-step explanation:

Given that:

$H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55$

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }$

$Z = \dfrac{-1}{\dfrac{5.3}{6} }$

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

b

For x = 53 and s = 4.6

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }$

$Z = \dfrac{-2}{\dfrac{4.6}{6} }$

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

$Conclusion: \ There \ is \ sufficient \ evidence \ to \ conclude \ that$ $\mu < 55$

c)

For x = 56 and s = 5.0

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

6 0

3 years ago

a fruit farm grows 4486 apple trees in 5 different Orchards each Orchard has about the same number of apples trees which is the

wariber [46]

Answer:

897.2

Steps:

1) 4486 divided by 5 = 897.2

2) Estimate:

We know that if the number is less than 5 the number in front of it stays the same, so the answer will be 897.

5 0

3 years ago

Your grade at the start of quarter 3 was a 68, at the end of the quarter it was a 91. What is the percent change in grade?

Georgia [21]

91 - 68 = 23
23/91 x 100= 25.3%

7 0

3 years ago