The mean life of a certain brand of auto batteries is 44 months with a standard deviation of 3 months. Assume that the lives of all auto batteries of this brand have a bell-shaped distribution. Using the empirical rule, find the percentage of auto batteries of this brand that have a life of 35 to 53 months. Round your answer to one decimal place.

99.7%

Step-by-step explanation:

Using the z score formula

z-score is z = (x-μ)/σ

where:

x = raw score

μ = population mean

σ = population standard deviation.

a) for x = raw score = 35

μ = population mean = 44

σ = population standard deviation =3

z = (35 - 44)/3

z = -9/3

z = -3

b) for x = raw score = 53

μ = population mean = 44

σ = population standard deviation =3

z = (53 - 44)/3

z = 9/3

z = 3

We would use the standard normal distribution table to find their probabilities

P(Z<=-3)=0.0013499

P(Z<=+3)=0.99865

So P(-3<=Z<=3)=0.99865 - 0.0013499 = 0.9973

Converting to percentage = 0.9973 × 100 = 99.73%

Therefore, the percentage of auto batteries of this brand that have a life of 35 to 53 months to one decimal place is 99.7%

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.