An equation represents a line that passes through (-2, 4) and has a slope of 2/5 is y-4 = 2/5(x+2)
<h3>Equation of a line</h3>
The equation of a line in point-slope form is expressed as;
y-y1 = m(x - x1)
where
m is the slope
(x1, y1) is any pint on the line
Substitute the given parameter
y - 4 = 2/5(x-(-2))
y-4 = 2/5(x+2)
Hence an equation represents a line that passes through (-2, 4) and has a slope of 2/5 is y-4 = 2/5(x+2)
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Answer:
Step-by-step explanation:
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:

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
Answer:
Question 1: Option C, 2x^2(x - 3)(x^2 + 3x + 9)
Question 2: Options 1, 2, 5
Step-by-step explanation:
Question #3
Step 1: Factor
2x^5 − 54x^2
2x^2(x^3 - 27)
<em>2x^2(x - 3)(x^2 + 3x + 9)</em>
<em />
Answer: Option C, 2x^2(x - 3)(x^2 + 3x + 9)
Question #2
p(x) = 4x^6 + 32x^3
<u>Step 1: Factor</u>
4x^6 + 32x^3
4x^3(x^3 + 8)
<em>4x^3(x + 2)(x^2 - 2x + 4)</em>
<em />
Answer: Options 1, 2, 5