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Leto [7]
3 years ago
10

Please help X-2 If = k and k = -4, what is the value of x? 5

Mathematics
1 answer:
SashulF [63]3 years ago
7 0

Answer:

C

Step-by-step explanation:

So we have the equation:

\frac{x-2}{5}=k

And we want to find the value of x such that k is -4.

To do so, we can substitute -4 for k and then solve for x.

So, substitute -4 for k:

\frac{x-2}{5}=-4

And now, we just have to solve for x.

Let's multiply both sides by 5:

(5)\frac{x-2}{5}=5(-4)

The left side cancels. Multiply on the right:

x-2=-20

Now, let's add 2 to both sides:

(x-2)+2=(-20)+2

The left side cancels again. Add on the right:

x=-18

Therefore, the value of x is -18.

Our answer is C.

And we're done!

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What number is composed of a whole number and a fraction
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Answer:

When an expression consists of a whole number and a proper fraction, we define it as a mixed number.

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3 years ago
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.

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is the answer
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