A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 39 months and a standard deviation of 10 months. Using the 68-95-99.7 rule, what is the approximate percentage of cars that remain in service between 59 and 69 months? Do not enter the percent symbol.

2.35%

Step-by-step explanation:

Mean number of months (M) = 39 months

Standard deviation (S) = 10 months

According to the 68-95-99.7 rule, 95% of the data is comprised within two standard deviations of the mean (39-20 to 39+20 months), while 99.7% of the data is comprised within two standard deviations of the mean (39-30 to 39+30 months).

Therefore, the percentage of cars still in service from 59 to 69 months is:

The approximate percentage of cars that remain in service between 59 and 69 months is 2.35%.

they disappear if your absent from to long. i found out that the hard way.

step-by-step explanation:

brainliest lol i just need one more brainliest.

(a) common ratio in the given sequence is, 3

(b) push ups will do kendall do on the 20th week

step-by-step explanation:

geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio(r)

(a)

given the sequence: 6 , 18 , 54 , 162 ,

this sequence is geometric sequence where each term is 3 times the previous term.

therefore,

common ratio(r) = 3.

(b)

a formula for nth term of geometric sequence is given by;

,   where r is the common ratio between the successive terms and is the first term,

to find how many push ups will kendall do on the 20th week.

here. n = 20th term.

in the given sequence first term

then;

simplify:

therefore, push ups will do kendall do on the 20th week