Arandom sample of 300 americans planning long summer vacations in 2009 revealed an average planned expenditure of $1,076. the expenditure for all americans planning long summer vacations has a normal distribution with a standard deviation 345. give a 99% confidence interval for the mean planned expenditure by all americans taking long summer vacations in 2009. explain your answer in relation to this context.

Step-by-step explanation:

We know that the sample size was:

The average was:

The standard deviation was:

The confidence level is

The confidence interval for the mean is:

Looking at the normal table we have to

Therefore the confidence interval for the mean is:

This means that the mean planned spending of all Americans who take long summer vacations in 2009 is between $ 1024.69 and $ 1127.31

step-by-step explanation:

4227473658779875563886538705958641728563470543621287463824132573847969p05749365273847906970680574936214

y = 5x + 17

step-by-step explanation:

a perpendicular line will be the opposite reciprocal slope of the given equation.   the slope given is -1/5 so the opposite reciprocal is 5

find b:

7 = 5 (-2) + b

7 = -10 + b

17 = b

write equation

y = 5x + 17