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Likurg_2 [28]
3 years ago
12

Write the point-slope form of the equation of the line through the given points.

Mathematics
1 answer:
insens350 [35]3 years ago
8 0

Answer:

y-3=7/6(x-4)

Step-by-step explanation:

the equation for point-slope form is y-y1=m(x-x1)

first let's find the slope (m)

the formula for slope is (y2-y1)/(x2-x1)

the given points: (4,3) and (-2,-4)

label the points:

x1=4

y1=3

x2=-2

y2=-4

now substitute into the equation

m=(-4-3)/(-2-4)

m=-7/-6

m=7/6

the slope is 7/6

now substitute the numbers into the equation for point-slope form

y-3=7/6(x-4)

Hope this helps :)!

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3 years ago
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
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Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
3 years ago
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