A police radar gun is used to measure the speeds of cars on a highway. The speeds of cars are normally distributed with a mean of 55 mi/hr and a standard deviation of 5 mi/hr. Roughly what percentage of cars are driving less than 45 mi/hr? (Round to the nearest tenth of a percent)

the percentage of cars that are driving less than 45 mi/hr is 2.3%

Step-by-step explanation:

Since the speeds of cars are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = speeds of cars

µ = mean speed

σ = standard deviation

From the information given,

µ = 55 mi/hr

σ = 5 mi/hr

The probability that a car is driving less than 45 mi/hr is expressed as

P(x < 45)

For x = 45

z = (45 - 55)/5 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.023

Therefore, the percentage of cars that are driving less than 45 mi/hr is

0.023 × 100 = 2.3%

answer: 2x + 2 − 2 = 10 − 2; subtraction property of equality

step-by-step explanation:

f . i can go whenever i feel like it bc it is confidential meaning only i would know how much i deposited besides the bank and they wouldnt care bc when you make a deposit you are using their services which means they are using there competition and gaining more customers

and if they dont well then thats less customers for them in a less profitable way

plz mark brainliest