Nomas necesito la respuesta para (D)

Is this expression a perfect square trinomial?

Katyanochek1 [597]

Yes it is (10x-y)^2 is the answer of it

6 0

4 years ago

Solve the system using substitution.<br> 2x = 5y + 14<br> x = 5y

andrew-mc [135]

Answer:

Step-by-step explanation:

To solve an equation, we can only have one variable. In this case, we are given the system 2x = 5y + 14. However, we are given the value of x in terms of y. We can substitute 10y in for 2x and solve.

10y = 5y + 14

First, we need to get all the variables on one side of the equation, so we subtract 5y.

5y = 14

Divide by 5

y = 14/5 or 2 4/5 or 2.8

Now for the x part.

2x = x + 14

Subtract 1x

x = 14

Now we check our answer.

2(14) = 5(2.8) + 14

28 = 14 + 14

28 = 28

Because 28 equals 28, our solutions are correct.

7 0

3 years ago

In a sample of 42 burritos, we found a sample mean 1.4 lb and assumed that sigma equals.5. In a test of the hypothesis H subscri

Sergio [31]

Answer:

D. 0.9953 (Probability of a Type II error), 0.0047 (Power of the test)

Step-by-step explanation:

Let's first remember that a Type II error is to NOT reject H0 when it is false, and the probability of that occurring is known as β. On the other hand, power refers to the probability of rejecting H0 when it is false, so it can be calculated as 1 - β.

To resolve this we are going to use the Z-statistic:

Z = (X¯ - μ0) / (σ/√n)

where μ0 = 1.2

σ = 5

n = 42

As we can see in part A of the attached image, we have the normal distribution curve representation for this test, and because this is a two-tailed test, we split the significance level of α=0.01 evenly into the two tails, 0.005 in each tail, and if we look for the Z critical value for those values in a standard distribution Z table we will find that that value is 2.576.

Now we need to stablish the equation that will telll us for what values of X¯ will we reject H0.

Reject if:

Z ≤ -2.576 Z ≥ 2.576

We know the equation for the Z-statistic, so we can substitute like follows and resolve.

Reject if:

(X¯ - 1.2) / (5/√42) ≤ -2.576 (X¯ - 1.2) / (5/√42) ≥ 2.576

X¯ ≤ -0.79 X¯ ≥ 3.19

We have the information that the true population mean is 1.25, so we now for a fact that H0 is false, so with this we can calculate the probability of a Type II error: P(Do not reject H0 | μ=1.25)

As we can see in part B of the attached image, we can stablish that the type II error will represent the probability of the sample mean (X¯) falling between -0.79 and 3.19 when μ=1.25, and that represents the shaded area. So now we now that we are looking for P(-0.79 < X¯ < 3.19 | μ=1.25).

Because we know the equation of Z, we are going to standardize this as follows:

P ( (-0.79 - 1.25) / (5/√42) < Z < (3.19 - 1.25) / (5/√42) )

This equals to:

P(-2.64 < Z < 2.51)

If we go and look for the area under the curve for Z positive scores in a normal standart table (part C of attached image), we will find that that area is 0.9940, which represents the probability of a Type II error.

Therefore, the power of the test will be 1-0.9940 = 0.006

If we look at the options of answers we have, there is no option that looks like this results, which means there was a probable redaction error, so we are going to stay with the closest option to these values which is option D.