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3 years ago

14

Line AB passes through points A(–6, 6) and B(12, 3). If the equation of the line is written in slope-intercept form, y = mx + b,

then m = – and b = .

1 answer:

inn [45]3 years ago

8 0

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points A(-6, 6) and B(12, 3). Substitute:

$m=\dfrac{3-6}{12-(-6)}=\dfrac{-3}{18}=-\dfrac{3:3}{18:3}=-\dfrac{1}{6}$

Therefore we have $y=-\dfrac{1}{6}x+b$ .

Put the coordinates of the point B to the equation:

$3=-\dfrac{1}{6}(12)+b$

$3=-2+b$ <em>add 2 to both sides</em>

$5=b\to b=5$

Answer:

$m=-\dfrac{1}{6}\\\\b=5$

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a) 0.5.

b) 0.8413

c) 0.8413

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f) 1

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

$Z = \frac{6-6}{0.2}$

$Z = 0$

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(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

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$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

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

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This is 1 subtracted by the pvalue of Z when X = 5.7.

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This is 1 subtracted by the pvalue of Z when X = 5.

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