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skelet666 [1.2K]
3 years ago
14

A shirt originally cost $51.06, but it is on sale for $35.74. What is the percentage decrease of the price of the shirt? If nece

ssary, round to the nearest percent.
24%





30%





43%





70%
Mathematics
1 answer:
KIM [24]3 years ago
8 0
The answer is the third choice <span />
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I think it is a and f i could be wrong

Step-by-step explanation:

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Car A travels 18 miles every 20 minutes. Car B travels 13 miles every 12 minutes. Which Statement best compares how far the two
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Car A: 18/20(60)=54 mph

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So car b travels 11 more miles than car a in one hour

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Find the slope of the line.
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rise over run

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If f(x)=7x-5, find f-|(16) ?<br>​
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A national manufacturer of ball bearings is experimenting with two different processes for producing precision ball bearings. It
lapo4ka [179]

Answer:

A. As we are trying to test if the two processes yield different average errors, we are interested in any of the two posibilities: the average error of process A being statistically significant lower or bigger than the average error of Process B. That is why this is a two-tailed test.

B. t=-3.144

C. The critical value for a significance level of 0.05, a two tailed test with 4 degrees of freedom is t=±2.064.

D. P-value = 0.004

E. Reject H0

F. The probability of making a Type I error is equal to the significance level: P(Type I error) = 0.05.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the two processes yield different average errors.

As we are trying to test if the two processes yield different average errors, we are interested in any of the two posibilities: the average error of process A being statistically significant lower or bigger than the average error of Process B. That is why this is a two-tailed test.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

being  μ1: average error for Process A and μ2: average error fo Process B.

The significance level is 0.05.

The sample 1, of size n1=12 has a mean of 2 and a standard deviation of 1.

The sample 1, of size n1=14 has a mean of 3 and a standard deviation of 0.5.

The difference between sample means is Md=-1.

M_d=M_1-M_2=2-3=-1

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1^2}{12}+\dfrac{0.5^2}{14}}\\\\\\s_{M_d}=\sqrt{0.083+0.018}=\sqrt{0.101}=0.318

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-1-0}{0.318}=\dfrac{-1}{0.318}=-3.144

The degrees of freedom for this test are:

df=n_1+n_2-1=12+14-2=24

The critical value for a significance level of 0.05, a two tailed test with 4 degrees of freedom is t=±2.064.

This test is a two-tailed test, with 24 degrees of freedom and t=-3.144, so the P-value for this test is calculated as (using a t-table):

P-value=2\cdot P(t

As the P-value (0.004) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the two processes yield different average errors.

5 0
3 years ago
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