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Talja [164]
3 years ago
5

N-9 = 13 ------------------

Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
5 0

Answer:

n-9=13

n=22

Step-by-step explanation:

Maslowich3 years ago
3 0

Answer: n = 22

Step-by-step explanation: to get the answer you have to add 9 to the -9 so that it cancels out. then you are to add the 9 to the 13 and so the answer shall be n=22

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Calculate the pay for the following day of a
DedPeter [7]

Answer:

Step-by-step explanation:

The person works 8.5 hours that day at a rate of $16/hr. 16 x 8.5 = $136

8 0
2 years ago
An established company started with 147 employees. After 10 years in business, they now have 283 employees. What is the percent
xeze [42]
So, Increase = New Number - Original Number. Therefore, Increase = 283 - 147 which equals 136.
136 = 283 - 147.
Then you want to divide the increase by the original number and multiply the answer by 100.
136/147 = 0.92
0.92 * 100 = 92.

Final Answer: The employee count has increased by 92%. I hope this helps.

4 0
3 years ago
PLEASE HELP<br> I NEED BOTH PARTS!
MAVERICK [17]
I think it’s b but don’t take my word for it
6 0
2 years ago
Help fast please ASAP
Aloiza [94]

Answer:

a rotation of 180° about point (0.5, -0.5) followed by a dilation of 0.5

Step-by-step explanation:

3 0
2 years ago
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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