Fast-food restaurants spend much time studying the amount of time cars spend in their drive-thrus. Certainly, the faster the cars get service, the more opportunity for making money. According to a recent study by QSR magazine, Wendy’s has the best time, with a mean time spent in the drive thru of 138.5 seconds. Assuming drive-thru time is normally distributed with a standard deviation of 29 seconds, what proportion of cars spends between 120 and 180 seconds in Wendy's drive-thru? Write answer as decimal rounded to the thousandth.

0.663

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

What proportion of cars spends between 120 and 180 seconds in Wendy's drive-thru?

This is the pvalue of Z when X = 180 subtracted by the pvalue of Z when X = 120. So

X = 180

has a pvalue of 0.924

X = 120

has a pvalue of 0.261

0.924 - 0.261 = 0.663

the y will be in still but your answer will be y x 3

step-by-step explanation:

jk: )

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.