The mileages of cars in a fleet of cars are normally distributed with a mean mileage of 52,000 miles and a standard deviation of 2,000 miles. What is the probability of randomly selecting a car with a mileage less than 52,000 miles?

50%

Step-by-step explanation:

We have that the mean is 52000 miles (m) and the standard deviation is 2000 miles (sd), so we are asked:

P (x <52000)

P ((x - m) / sd <(52000 -52000) /2)

P (z <0)

If we look in the normal distribution table we have to:

P (x <52000) = 0.5

That is, the probability is of randomly selecting a car with a mileage less than 52,000 miles 50%

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

answer: it's d

step-by-step explanation: