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aleksandr82 [10.1K]

2 years ago

14

What is 25% of 25,000

2 answers:

prisoha [69]2 years ago

6 0

You do cross multiplying so 25/100 because a 100 is the most of a percent you can get so it that percent over a 100=x/25,000. there's a x because you don't know what the number on top of the 25,000 is.Now do cross multiplying, 25 times 25,000 divided by 100 is 6250

Sunny_sXe [5.5K]2 years ago

6 0

25% of 25,000 is 6250

Convert the percentage into a decimal by dividing it over 100:
$\frac{25}{100} = 0.25$

Multiply the decimal by 25,000:
$0.25 \times 25000 = 6250$

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2 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

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Answer:

Step-by-step explanation:

Given that:

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

(a) For x = 54 and s = 5.3

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

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

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Using the Excel Formula:

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

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b

For x = 53 and s = 4.6

The test statistics can be computed as:

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

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

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

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degree of freedom = n - 1

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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:

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

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

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

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degree of freedom = n - 1

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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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