Students are traveling in two cars to a football game 135 miles away. the first car leaves on time and travels at an average speed of 45 miles per hour. the second car starts 1/2 hour later and travels at an average speed of 55 miles per hour. how long will it take the second car to catch up to the first car? will the second car catch up to the first car before the first car arrives at the game?

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)

i think da answer is

d.square root of 3

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.