When Easton goes bowling, his scores are normally distributed with a mean of 130 and a standard deviation of 11. What percentage of the games that Easton bowls does
he score higher than 112, to the nearest tenth?

P(X≥112) = 0.9495

The  percentage of the games that Easton bowls does  he score higher than 112 = 94.95%

Step-by-step explanation:

Given mean of the Population = 130

Given standard deviation of the Population = 11

Let X= 112

P(X≥112) = P(Z≥ z)

               = 1-P(Z≤z)

               = 1- ( P(Z≤-1.636)

             =  1- (0.5-A(-1.636)

            = 0.5 +A(1.636)

            = 0.5 + 0.4495

P(X≥112) = 0.9495

The  percentage of the games that Easton bowls does

he score higher than 112 = 94.95%

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2