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3 years ago

7

At her job, mary works 4 hours on monday, 3 hours on wednesday, and 6 hours on saturday. her boss adds a $10 bonus to her payche

ck, which then totals $120.50. what is mary's hourly pay?
a. $8.50
b. $9.27
c. $5.24
d. $8.05

1 answer:

Andru [333]3 years ago

8 0

She worked 4+3+6 hours = 13 hoursso 13x+10 = 120.50
13x=110.50
x=8.5
so she makes $8.50 an hour which is A.

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She had $3000. She spend 40% of the money in smart TV and 15% of the remainder on Game Set. (a) What % of the money had she left

Dovator [93]

Answer: She had left 54% of the money which is $1620.

Step-by-step explanation:

40% of $3000 = $1200

15% of $1200 = $180

$3000 - ($1200 + $180)

$3000 - $1380 = $1620

$1620 is the money she had left.

$1620 is 54% of $3000

She had 54% left of the money.

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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:

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

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

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

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 \$ $\mu < 55$

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3 years ago

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ZanzabumX [31]

The inequality that describes this graph of the number line is x ≥ 4

<h3>How to determine the inequality that describes this graph?</h3>

From the question, we have the following properties:

The graph is a number line
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The number line starts from less than -5 and ends at greater than 5
There is a heavy arrow that extends from a circle over positive 4 to the right.

The property 3 above implies that the domain of the number line is the set of all real numbers.

The property 4 above implies that the inequality of the number line is a greater than or equal to inequality, and the value starts from 4

This is represented by the following inequality

x ≥ 4

Hence, the inequality that describes this graph of the number line is x ≥ 4

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Answer:

Step-by-step explanation:

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