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Zolol [24]
3 years ago
15

Write, in slope intercept form, the equation of the line that passes through (8, -9) and has a slope of -5/4.

Mathematics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:

The equation in the slope-intercept form will be:

y=-\frac{5}{4}x+1

Step-by-step explanation:

Given

  • slope = m = -5/4
  • point = (8, -9)

As we know that the equation of a line in point-slope form is

y-y_1=m\left(x-x_1\right)

substituting the values m = -5/4 and point = (8, -9)

y-\left(-9\right)=\frac{-5}{4}\left(x-8\right)

y+9=\frac{-5}{4}\left(x-8\right)

Writing the equation in slope-intercept form

y=mx+b

where m is the slope, and b is the y-intercept

so the equation of the line in slope-intercept form becomes

y+9=\frac{-5}{4}\left(x-8\right)

subtract 9 from both sides

y+9-9=\frac{-5}{4}\left(x-8\right)-9

y=-\frac{5}{4}x+1

Therefore, the equation in the slope-intercept form will be:

y=-\frac{5}{4}x+1

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A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc
pogonyaev

Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

(a) What are the mean and the standard deviation of the number X of students who accept?

n = 1200, p = 0.75. So

E(X) = np = 1200*0.75 = 900

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15

The mean is 900 and the standard deviation is 15.

(b) Use the Normal approximation to find the probability that at least 800 students accept.

Using continuity corrections, this is P(X \geq 800 - 0.5) = P(X \geq 799.5), which is 1 subtracted by the pvalue of Z when X = 799.5. So

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

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(c) The college does not want more than 950 students. What is the probability that more than 950 will accept?

Using continuity corrections, this is P(X \geq 950 - 0.5) = P(X \geq 949.5), which is 1 subtracted by the pvalue of Z when X = 949.5. So

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

Z = \frac{949.5 - 900}{15}

Z = 3.3

Z = 3.3 has a pvalue of 0.9995

1 - 0.9995 = 0.0005

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(d) If the college decides to increase the number of admission offers to 1300, what is the probability that more than 950 will accept?

Now n = 1300. So

E(X) = np = 1300*0.75 = 975

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15.6

Same logic as c.

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

Z = \frac{949.5 - 975}{15.6}

Z = -1.63

Z = -1.63 has a pvalue of 0.0516

1 - 0.0516 = 0.9484

94.84% probability that more than 950 will accept

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