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Ne4ueva [31]
3 years ago
6

The result of adding a complex number to its conjugate is “an integer/a pure imaginary number/a real number/a whole number” and

the result of subtracting a complex number from its conjugate is “an integer/a real number/a pure imaginary number/a whole number”
Mathematics
1 answer:
sineoko [7]3 years ago
6 0

Answer:

If Z is a complex number:

Z = a + b*i

where a and b are real numbers, and i is an imaginary number.

Then "a" is the real part.

"b*i" is the imaginary part.

The conjugate of Z is:

Zc = a - b*i

So the sign of the imaginary part changes.

Then:

Sum:

Z + Zc = (a + bi) + (a - bi) = 2*a + 0 = 2*a

and remember that a is a real number, then 2*a is also a real numer.

The correct answer is "A real number".

Difference:

Z - Zc = (a + bi) - (a - bi) = 2b*i

and this is a pure imaginary number, so here the correct answer is: "a pure imaginary number"

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1. slope of the given line = 1/5
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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.

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melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

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<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

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0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

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-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

-2x\:

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)>10\left(-1\right)

Simplify

2x>-10

Divide both sides by 2

\frac{2x}{2}>\frac{-10}{2}

x>-5

-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

-2x>-6

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)

Simplify

2x

Divide both sides by 2

\frac{2x}{2}

x

-6

Thus, the two intervals:

\left(-\infty \:,\:3\right)

\left(-5,\:\infty \:\right)

The intersection of these two intervals would be the solution to both inequalities.

\left(-\infty \:,\:3\right)  and \left(-5,\:\infty \:\right)

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR  -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0
3 years ago
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