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PIT_PIT [208]
2 years ago
14

John must mount a box which is 12 5/8 inches wide. He wants to leave 7 1/2 inches on each side of the

Mathematics
2 answers:
AveGali [126]2 years ago
5 0

Answer:

27 5/8

Step-by-step explanation:

12 5/8 + 2(7 1/2) = 12 5/8 + 15 = 27 5/8

Veseljchak [2.6K]2 years ago
4 0

Answer:

the answer is

 2 * 7 1/2 = 15 12 5/8 + 15 = 27 5/8

Step-by-step explanation:

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The median average refers to number in the middle of the set. But we need to write it in chronological order.

Now the set becomes 57, 60, 62, 65.

Now find the number in the middle. In this case, it both 60 and 62.
But we need one number. 

So, we will take the average of both of them.

60 + 62 = 122

122 / 2 = 61

So, 61 is the median of the set.

EDIT:

Big thanks to mkersten for finding two errors in my answer!
7 0
3 years ago
If the mean of a given dataset is 131 and the standard deviation is 8, where will a majority of the data lie?
Anika [276]

Answer:

A

Step-by-step explanation:

The reason it has to be A is because you don't have to even thing about the standard deviation right here because the only answer that has a mean of 131 is answer A.


Hope it helps have a great afternoon:)

8 0
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3 years ago
I need help bad I’m struggling
Likurg_2 [28]

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3 years ago
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A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st
ss7ja [257]

Answer:

Step-by-step explanation:

Hello!

The researcher wants to test if the valve plates manufactured have the expected tensile strength of 5 lbs/mm. So he took a sample of 42 valve plates and measured their tensile strength, obtaining a sample mean of X[bar]= 5.0611 lbs/mm and a sample standard deviation of S=0.2803 lbs/mm.

The study variable is:

X: tensile strength of a valve plate (lbs/mm)

The parameter of interest is the mean tensile strength of the valve plates, μ.

If the claim is that the valve plates of the sample have on average tensile strength of 5 lbs/mm, symbolically: μ = 5

a) The statistic hypotheses are:

H₀: μ = 5

H₁: μ ≠ 5

b) To determine the critical values and rejection region of a hypothesis test you need three to determine three factors of the hypothesis test:

1) The statistical hypothesis.

2) The significance level.

3) The statistic to use for the analysis.

The statistic hypothesis determines the number of critical values and the direction of the rejection region, in this case, the test is two-tailed you will have two critical values and the rejection region will be divided into two.

With the statistic, you will determine the distribution under which you will work and the significance level determines the probability of rejecting the null hypothesis.

To study the population mean you need that the variable of interest has at least a normal distribution, there is no information about the distribution of the study variable but the sample size is large enough n≥30, so you can apply the central limit theorem to approximate the distribution of the sample mean to normal: X[bar]≈N(μ;σ²/n)

Thanks to this approximation it is valid to use an approximation of the standard normal distribution for the test:

Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }≈N(0;1)

The critical values are:

Z_{\alpha /2}= Z_{0.05}= -1.648

Z_{1-\alpha /2}= Z_{0.95}= 1.648

You will reject the null hypothesis if Z_{H_0}≤-1.648 or if Z_{H_0}≥1.648

You will not reject the null hypothesis if -1.648<Z_{H_0}<1.648

c)

Z_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }=  \frac{5.0611-5}{\frac{0.2803}{\sqrt{42} } }= 1.41

d) The value of the statistic is between the two critical values so the decision is to not reject the null hypothesis. Then using a significance level of 10% there is no significant evidence to reject the null hypothesis so the valve plates have on average tensile strength of 5 lbs/mm.

e) The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis). If the test is two-tailed, so is the p-value, you can calculate it as:

P(Z≤-1.41) + P(Z≥1.41)= P(Z≤-1.41) + (1 - P(Z≤1.41))= 0.079 + ( 1 - 0.921)= 0.158

p-value: 0.158

I hope it helps!

5 0
3 years ago
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