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3 years ago

The cost of four sweaters an two pairs of jeans is 140$. The cost of two sweaters an three pairs of jeans is $150.

Mathematics

2 answers:

bearhunter [10]3 years ago

7 0

sweater cost $20 and jeans cost $30

Kisachek [45]3 years ago

3 0

Answer:

A pair of jeans cost 40 dollars

A sweater costs 15 dollars

Step-by-step explanation:

Let s = cost a of sweater

j = cost of a pair of jeans

4s +2j = 140

2s +3j = 150

I will use elimination to solve. Multiply the second equation by -2

-2(2s +3j )= 150*-2

-4s -6j = -300

Add this equation to the first equation

-4s -6j = -300

4s +2j = 140

--------------------

-4j = -160

Divide by -4

-4j/-4 = -160/-4

j = 40

A pair of jeans cost 40 dollars

Now we need to find the cost of the sweater

2s +3j = 150

2s +3(40) = 150

2s+120 = 150

Subtract 120 from each side

2s +120-120=150-120

2s = 30

Divide by 2

2s/2=30/2

s = 15

A sweater costs 15 dollars

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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 \$ $\mu < 55$

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.

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

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 \$ $\mu < 55$

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