The mean of normally distributed test scores is 82 and the standard deviation is 5. If there are 241 test scores in the data sample, how many of them were in the 92 to 97 range?

5

Step-by-step explanation:

Find the z-scores.

z = (x − μ) / σ

z₁ = (92 − 82) / 5

z₁ = 2

z₁ = (97 − 82) / 5

z₂ = 3

Find the probability:

P(92 < X < 97)

P(2 < Z < 3)

P(Z < 3) − P(Z < 2)

0.9987 − 0.9772

0.0215

Find the number of tests:

0.0215 (241) ≈ 5

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form