Answer:
d
Step-by-step explanation:
Answer:
A) 1
B) 1
C) 0
Step-by-step explanation:
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n = 49 customers is observed. Find the probability that the average time waiting in line for these customers is A) Less than 10 minutes B) Between 5 and 10 minutes C) Less than 6 minutes
We solve this question using the z score formula
z = (x-μ)/σ/√n, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
n = number of random samples
A) Less than 10 minutes
x < 10
z = 10 - 8.2/ 1.5 / √49
z = 8.4
P-value from Z-Table:
P(x<10) = 1
B) Between 5 and 10 minutes
For x = 5 minutes
z = 5 - 8.2/ 1.5 / √49
z = -14.93333
P-value from Z-Table:
P(x = 5) = 0
For x = 10 minutes
z = 10 - 8.2/ 1.5 / √49
z = 8.4
P-value from Z-Table:
P(x = 10) = 1
The probability that the average time waiting in line for these customers is between 5 and 10 minutes
P(x = 10) - P(x = 5)
= 1 - 0
= 1
C) Less than 6 minutes
x < 6
z = 6 - 8.2/ 1.5 / √49
z = -10.26667
P-value from Z-Table:
P(x<6) = 0
Answer:
The terms in #4 are 2500, 11t, and -3w
The terms in #5 are 17 and 4d
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
First take 18 and divide by 50 = .36, so your percent is 36%.
