If there were 49 cars in a line that stretched 452 feet, what is the average car length? assume that the cars are lined up bumper-to-bumper. round your answer to the nearest tenth of a foot.

The average car length = 9.2 ft.

Step-by-step explanation:

Here, the number of cars lined up = 49

The total stretched length of the cars  = 452 ft.

Let us assume the length of 1 car  = p ft.

So, the length of 49 cars combined  = 49 x ( Length of 1 car)

                                                             = 49 x ( p)  = 49 p

Also, the combined length is given as 452 ft.

Now, rounding off the answer to nearest tenth of a foot, we get:

p =  9.2 ft.

Hence, the average car length = 9.2 ft.

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497