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larisa86 [58]
3 years ago
12

I really need to do #3

Mathematics
1 answer:
Aloiza [94]3 years ago
8 0
A.20%
Because if we change 20% to 100% , you will need to multiply it by 5. The difference of 20 and 16 is 4, and 4×5=20
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Answer:

Step-by-step explanation:

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4(x^3+8)

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The equation 2x^2 + x - 1 = 0 has two solutions. Find an equation of the form ax^2 + bx + c = 0, which solutions....
leonid [27]

Answer:

Step-by-step explanation:

Let the solution to

2x^2 + x -1 =0

x^2+ (1/2)x -(1/2)

are a and b

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A. We want to have an equation where the roots are a +5 and b+5.

Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.

The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.

So the equation is

x^2-(19/2)x + 22 =0

2x^2-19x + 44 =0

B. We want the roots to be 3a and 3b.

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(3a)(3b) = 9(ab) =9(-1/2)=-9/2.

So the equation is

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In 2014, the CDC estimated that the mean height for adult women in the U.S. was 64 inches with a standard deviation of 4 inches.
Vladimir79 [104]

Answer:

A. 16%

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 64, \sigma = 4

Which of the following gives the probability that a randomly selected woman has a height of greater than 68 inches?

This is 1 subtracted by the pvalue of Z when X = 68. So

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

Z = \frac{68 - 64}{4}

Z = 1

Z = 1 has a pvalue of 0.84.

1 - 0.84 = 0.16

So the correct answer is:

A. 16%

6 0
3 years ago
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