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Delvig [45]
2 years ago
7

Write a linear equation given the point and slope or two points.

Mathematics
1 answer:
attashe74 [19]2 years ago
5 0

Answer:

1) y = -2x - 1

2) y = -3/4 + 3

3) y = 4x + 9

4) y = - 5/3x - 2

Step-by-step explanation:

b = y - m*x

1) (-7,13) and slope: -2

b = 13 - (-2)*(-7)

b = 13 - 14

b = -1

2) (4,6) lope = -3/4

b = 6 - (-3/4)*(-4)

b = 6 - 3

b = 3

y = -3/4x + 3

3) (-5,-11) and (3,-7)

Slope: (1 - - 11)/(-2 - - 5) = 12/3

= 4

b = -11 - (4) (-5) = 9

b = 9

4) slope = (-7 - 8)/(3- - 6)

= -15/9 = - 5/3

b = 8 - (-5/3)*(-6)

b = 8 - 10

b = -2

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Answer:

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The average midterm score of students in a certain course is 70 points.

This means that \mu = 70

29 students are randomly selected and the standard deviation of their scores is found to be 13.15 points.

This means that \sigma = 13.15, n = 29, s = \frac{13.15}{\sqrt{29}} = 2.44

Find the probability that the average midterm score of these students is at most 75 points.

This is the pvalue of Z when X = 75. So

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

By the Central Limit Theorem

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

Z = \frac{75 - 70}{2.44}

Z = 2.05

Z = 2.05 has a pvalue of 0.98.

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

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pogonyaev

Answer:

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Step-by-step explanation:

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E
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Answer:

1) n = 84

2) n = 78

3) n = 24

4) n = 16

5) n = 35

Step-by-step explanation:

To solve each proportion, we apply cross multiplication.

Question 1:

\frac{5}{12} = \frac{35}{n}

5n = 35*12

Simplifying both sides by 5

n = 7*12 = 84

Question 2:

\frac{n}{52} = \frac{180}{120}

120n = 52*180

Simplifying both sides by 20

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Simplifying by 3

2n = 52*3

Simplifying by 2

n = 26*3 = 78

Question 3:

\frac{18}{n} = \frac{21}{28}

21n = 18*28

Simplifying both sides by 7

3n = 18*4

Simplifying both sides by 3

n = 6*4 = 24

Question 4:

\frac{n}{4} = \frac{24}{6}

\frac{n}{4} = 4

n = 16

Question 5:

\frac{10}{16} = \frac{n}{56}

16n = 56*10

Simplifying by 2, both sides

8n = 56*5

Simplifying by 8, both sides

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