Complete question:
He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is
a) less than 8 minutes
b) between 8 and 9 minutes
c) less than 7.5 minutes
Answer:
a) 0.0708
b) 0.9291
c) 0.0000
Step-by-step explanation:
Given:
n = 47
u = 8.3 mins
s.d = 1.4 mins
a) Less than 8 minutes:

P(X' < 8) = P(Z< - 1.47)
Using the normal distribution table:
NORMSDIST(-1.47)
= 0.0708
b) between 8 and 9 minutes:
P(8< X' <9) =![[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20%5B%5Cfrac%7B8-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%3C%20%5Cfrac%7BX%27-u%7D%7Bs.d%2F%20%5Csqrt%7Bn%7D%7D%20%3C%20%5Cfrac%7B9-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D)
= P(-1.47 <Z< 6.366)
= P( Z< 6.366) - P(Z< -1.47)
Using normal distribution table,

0.9999 - 0.0708
= 0.9291
c) Less than 7.5 minutes:
P(X'<7.5) = ![P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20P%20%5BZ%3C%20%5Cfrac%7B7.5-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D%20)
P(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
Step-by-step explanation:
Since , internal sum of angle of triangle is 180°
So,
66 + x +49 + x+83 = 180
or, 198 + 2x = 180
or, 2x = 180 -198
or, 2x = -18
hence, x = - 9
Now angle A = x + 49
= (-9) + 49
= 40
<u>Hence </u><u>measure</u><u> </u><u>of </u><u>angle </u><u>A </u><u>is </u><u>4</u><u>0</u><u>°</u><u>.</u>
Answer:
27850
Step-by-step explanation:
9/100 = x/30.95
9 x 30.95 = 278.5
278.5 x 100
In a regular polygon, each exterior angle equals each central angle.
central angle = 360 / #sides
#sides = 360 / central angle
#sides = 360 / 60
#sides = 6
Answer:
Fluffy's tail is 1/12 of a meter longer than Fireball's tail.
Step-by-step explanation:
1 1/3 - 1 1/4 = x
get common denominator (12)
1 4/12 - 1 3/12 = 1/12