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Ne4ueva [31]
3 years ago
7

A set of equations is given below: Equation C: y = 6x + 9 Equation D: y = 6x + 2 Which of the following best describes the numbe

r of solutions to the given set of equations? One solution No solution Two solutions Many solutions
Mathematics
2 answers:
7nadin3 [17]3 years ago
7 0
So we can get from these two equations that y = 6x + 2 = 6x + 9

6x + 2 = 6x + 9
0 = 7 , but that's incorrect, so this set has no solution.
katrin2010 [14]3 years ago
3 0
Hello again Kristian

Y = 6x + 9
Y = 6x + 2

We need to solve 6x + 9 for y in y = 6x + 2
Y = 6x + 2
6x + 9 = 6x + 2
6x - 6x = 2 - 9
0 = -7

No solutions

The correct option is the second option.

Good luck with your education!
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The distribution of heights for adult men in a certain population is approximately normal with mean 70 inches and standard devia
KengaRu [80]

Answer:

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]

Step-by-step explanation:

I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}

The values of the cumulative distribution function of W, denoted by \phi , can be found in the attached file. Lets call y = \frac{z-70}{4} . We have

0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)

Thus

\phi(y) = 1-0.1 = 0.9

by looking at the table, we find that y = 1.28, therefore

\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .

Download pdf
8 0
2 years ago
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