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2 years ago

What is the contrapositive of the following conditional?

Mathematics

1 answer:

Nimfa-mama [501]2 years ago

3 0

Answer:

To form the contrapositive of the conditional statement, interchange the hypothesis and the conclusion of the inverse statement. The contrapositive of "If it rains, then they cancel school" is "If they do not cancel school, then it does not rain."

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Step-by-step explanation:

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If one root is the square root of the other in the equation 8 + mx + 27x^2 = 0, find the value of m

ICE Princess25 [194]

Answer:

Step-by-step explanation:

6 0

2 years ago

I need graph help plz

Likurg_2 [28]

Answer:

Step-by-step explanation:

Part A

x-intercepts of the graph → x = 0, 6

Maximum value of the graph → f(x) = 120

Part B

Increasing in the interval → 0 ≤ x ≤ 3

Decreasing in the interval → 3 < x ≤ 6

As the price of goods increase in the interval [0, 3], profit increases.

But in the price interval of (3, 6] profit of the company decreases.

Part C

Average rate of change of a function 'f' in the interval of x = a and x = b is given by,

Average rate of change = $\frac{f(b)-f(a)}{b-a}$

Therefore, average rate of change of the function in the interval x = 1 and x = 3 will be,

Average rate of change = $\frac{f(3)-f(1)}{3-1}$

= $\frac{120-60}{3-1}$

= 30

6 0

2 years ago

the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?

miss Akunina [59]

The equation of a circle in standard form is

$(x - h)^2 + (y - k)^2 = r^2$

where (h, k) is the center of the circle, and r is the radius if the circle.

We need to find the radius and center of the circle.
We are given a diameter, so to find the center, we need the midpoint of the diameter.

M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)
The center is (0, 2).

To find the radius, we find the length of the given diameter and divided by 2.

$d = \sqrt{(-6 - 6)^2 + (6 - (-2))^2)}$

$d = \sqrt{144 + 64}$

$d = \sqrt{208}$

$r = \dfrac{d}{2} = \dfrac{\sqrt{208}}{2} = \dfrac{\sqrt{208}}{\sqrt{4}} = \sqrt{52}$

$(x - 0)^2 + (y - 2)^2 = (\sqrt{52})^2$

$x^2 + (y - 2)^2 = 52$

7 0

3 years ago

What is the answer to <br> 8(x-3)+7=2x(4-17)

nadezda [96]

Answer:

x=1/2

Step-by-step explanation:

First you distribute the 8 over the (x-3) giving you 8x-24. Do the same for 2x and (4-17). That should give you 8x-34x.

Your equation should look like : 8x-24+7=8x-34x

Next try to balance out your equation. You want your x's all by themselves in order to solve.

The 8x's cancel eachother out so you have -24+7= -34x.

Add the like terms of -24 and 7 to get -17.

Equation: -17= -34x

Divide both sides by -34 to get x by itself.

-17/-34 = x

Reduce by dividing the numerator and denominator by 17 to get -1/-2

Since they are both negative the fraction turns positive.

So the answer should be x=1/2

5 0

2 years ago

g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distrib

algol13

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .