Annual snowfall in Anchorage, Alaska, is normally distributed with a mean height of 76 inches and a standard deviation of 4.7 inches. What is the probability this year that it will snow between 71.4 and 80.7 inches in Anchorage? 67.75% 54.71% 38.14% 23.22%

67.75%

Step-by-step explanation:

Given:

Given that:

µ = 76 ; σ = 4.7

P(x < 80.7) - P(x < 71.4)

Obtain the standardized score, Z ; x = 71. 4

Zscore = (x - μ) / σ

P(x < 71.4) = (71.4 - 76) / 4.7

P(x < 71.4) = - 4.6 / 4.7

P(x < 71.4) = - 0.9787

P(z < 0.9787) = 0.16386

x = 80.7

P(x < 80.7) = (80.7 - 76) / 4.7

P(x < 80.7) = 4.7 / 4.7

P(x < 80.7) = 1

P(z < 1) = 0.84134

0.84134 - 0.16386 = 0.67748 = 67.748% = 67.75%

43

step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18