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Tanzania [10]
3 years ago
11

josh has a job mowing lawns. He charges$25 for each yard. He needs at least $400 for the new game system that he wants. write an

d solve an inequality to find out how many yards he must mow to make at least $400
Mathematics
1 answer:
Murljashka [212]3 years ago
5 0

25x\geq400

x\geq16

Hope Your Thanksgiving Went Well, Here's Some Help Cleaning Up

-TheKoolKid1O1

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A newspaper published an article about a study in which researchers subjected laboratory gloves to stress. Among 225 vinyl? glov
romanna [79]

Answer:

a)

H₀: ρ₁ ≤ ρ₂

H₁: ρ₁ > ρ₂

b)

Z= \frac{('p_{1} - 'p_{2}) - (p_{1} - p_{2})}{\sqrt{'p(1-'p)[\frac{1}{n_1} + \frac{1}{n_2} ]} } ≈ N(0;1)

c)

p-value < .00001

d)

Reject the null hypothesis.

The population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.

Step-by-step explanation:

Hello!

The objective is to test if the population proportion of vinyl gloves that leak viruses (population 1) is greater then the population proportion of latex gloves that leak viruses (population 2).

Sample 1 (Vinyl)

n₁= 225

Sample proportion 'ρ₁= 0.60

Sample 2 (Latex)

n₂= 225

Sample proportion 'ρ₂= 0.09

Pooled proportion: 'ρ= (x₁+x₂)/(n₁+n₂)= ( 'ρ₁ + 'ρ₂)/2= (0.60 + 0.09)/2= 0.345 ≅0.35

The hypothesis is:

H₀: ρ₁ ≤ ρ₂

H₁: ρ₁ > ρ₂

α: 0.05

The statistic to use is a pooled Z

Z= \frac{('p_{1} - 'p_{2}) - (p_{1} - p_{2})}{\sqrt{'p(1-'p)[\frac{1}{n_1} + \frac{1}{n_2} ]} } ≈ N(0;1)

The critical region is one-tailed to the right (positive), the critical value is:

Z_{1-\alpha } = Z_{0.95} = 1.64

If Z ≥ 1.64, you will reject the null hypothesis.

If Z < 1.64, you will not reject the null hypothesis.

Z= \frac{(0.6 - 0.09) - (0}{\sqrt{0.35(1-0.35)[\frac{1}{225} + \frac{1}{225} ]} } = 11.34

Since the calculated Z- value is greater than the critical value, the decision is to reject the null hypothesis. The population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.

P-value approach:

p-value < .00001

If you compare with the level of signification, you reach the same decision.

I hope it helps!

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