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3 years ago

I gotta answer this but I don’t want to be wrong

Mathematics

2 answers:

hjlf3 years ago

8 0

Answer:

A does not belong, i´m pretty sure.

Step-by-step explanation:

viktelen [127]3 years ago

7 0

I am almost 100% sure that it’s A

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It is assumed that the mean weight of a Labrador retriever is 70 pounds. A breeder claims that the average weight of an adult ma

Sindrei [870]

Answer:

z(s) is in the acceptance region we accept H₀. We don´t have enough evidence to support the breeder´s claim

Step-by-step explanation:

We will test the breeder´s claim at 95% ( CI) or significance level

α = 5 % α = 0,05 α /2 = 0,025

Sample Information:

sample size n = 45

sample mean x = 72,5 pounds

Sample standard deviation s = 16,1

1.-Hypothesis Test:

Null Hypothesis H₀ x = 70

Alternative Hypothesis Hₐ x ≠ 70

Alternative hypothesis contains the information about what kind of test has to be developed ( in this case it will be a two-tail tets)

2.-z (c) is from z-table z(c) = 1,96

3.- z(s) = ( x - 70 ) / 16,1 / √45

z(s) = (72,5 -70 ) *√45 / 16,1

z(s) = 2,5 * 6,71 / 16,1

z(s) = 1,04

4.-Comparing z(s) and z(c)

z(s) < z(c)

Then z(s) is in the acceptance region we accept H₀. We don´t have enough evidence to support the breeder´s claim

7 0

3 years ago

Read 2 more answers

Can someone find the slopes of these two please?

GenaCL600 [577]

Answer:

you put the points into the slope formula and solve : (y2-y1)/(x2-x1)

1. (4 -1) / (-2 -3)

3 / -5

2. (-1 -(-2)) / (5 - 0)

(-1 +2) / 5

1/ 5

Step-by-step explanation:

i hope this helped :)

8 0

3 years ago

Read 2 more answers

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide

kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error of a (1 - α)% confidence interval for population mean (μ) is:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

σ = $60

MOE = $2

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

$=[\frac{1.96\times 60}{2}]^{2}$

$=3457.44\\\approx 3458$

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0

3 years ago

Please answer the attached question.

Nikitich [7]

Answer : fx + fh - f(x) / h

6 0

3 years ago

What value should be added to the expression to create a perfect square?

Lilit [14]

Answer is D.Don't take out the square

8 0

4 years ago