The time spent waiting in the line is approximately normally distributed. The mean waiting time is 55 minutes and the standard deviation of the waiting time is 33 minutes. Find the probability that a person will wait for more than 88 minutes. Round your answer to four decimal places.

Zero.

Step-by-step explanation:

The variable here is t where t is = time spent waiting in the line

The values for this variable and in this study, are normally distributed.

The mean waiting time = 55 minutes

The standard deviation = 33 minutes

What is the probability that t is greater than 88 minutes?

Now, the confidence interval about the mean for this study is:

[55-33], [55+33]  = [22, 88]

Therefore, all values of t fall within this lower and upper limit or interval.

If that is the case, and recalling also that the distribution of values is normal, then it is impossible for t to be greater than 88 minutes.

Hence, P[t>88] = 0

To 4 decimal places, that would be 0.0000

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

d

step-by-step explanation:

i used the pitagoras theoreme