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liraira [26]
3 years ago
10

Can someone pleaseeee help and if you’re correct i’ll give brainliest

Mathematics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

Can you give me the website to see the problem or something.

Step-by-step explanation:

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Y+7= -2/5(x-10). What is the standard form
jeka94

Answer:

y=-2/5x+3

Step-by-step explanation:

y+7=-2/5(x-10)

step 1:

-2/5(x-10)= -2/5x-4

step 2:

y+7=-2/5x-4+7

step 3:

y=-2/5x+3

5 0
3 years ago
What’s the answer ?
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Step-by-step explanation:

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Can you help me find the area of the trapezoids ​
DiKsa [7]

Answer:

See below.

Step-by-step explanation:

the one on the left area is 104

the one on the right area is 19.5

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Anything that has mass and takes up space
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3 years ago
A psychologist wants to see if a certain company has fair hiring practices in an industry where 60% of the workers are men and 4
makvit [3.9K]

Answer:

A) H_{0}: p=0.5 (At least half of the workers are women,fair)

B) H_{a}: p<0.5 (Less than half of the workers are women,unfair)

C) critical value of the test statistic is 1.64 (one tailed)

D) Test statistic is ≈ 0.29

E) Since 0.29<1.64, we fail to reject the null hypothesis. There is no significant evidence that the company has unfair hiring practices at 0.05 significance level.

Step-by-step explanation:

Let p be the proportion of women workers in the company. Null and alternative hypotheses are

H_{0}: p=0.5 (At least half of the workers are women,fair)

H_{a}: p<0.5 (Less than half of the workers are women,unfair)

Test statistic can be found using the equation:

z=\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } } where

  • p(s) is the sample proportion of women workers (\frac{55}{107} =0.514)
  • p is the proportion assumed under null hypothesis. (0.5)
  • N is the sample size (55+52=107)

Then z=\frac{0.514-0.5}{\sqrt{\frac{0.5*0.5}{107} } } ≈ 0.29

For alpha 0.05, critical value of the test statistic is 1.64 (one tailed)

Since 0.29<1.64, we fail to reject the null hypothesis. There is no significant evidence that the company has unfair hiring practices at 0.05 significance level.

4 0
3 years ago
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