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
Answer:
c
Step-by-step explanation:
because if you look at the line below it it shows 180 degrees because it is a straight line if that makes sense. if not just trust me.
Answer:
The measurement which Jane finds to be 10 meters is the length of the banner.
Step-by-step explanation:
The measurement of 10 meters which Jane found after measuring how long the banner is before painting is the LENGTH of the banner.
This is clear from the unit of what she finds (meters). It only indicates the measurement of one part of the banner, even though a banner has two parts, the length and width.
It is possible to find the AREA, or PERIMETER, or LENGTH.
But what she finds is the LENGTH of the banner. If it was Area or Perimeter, the unit would have been square meters.
The distance between the two points (-1, 4) and (5, 4) is 6 units.
<u>Step-by-step explanation:</u>
The given Coordinate points are (-1, 4) and (5, 4).
The points are considered as (x1,y1) and (x2,y2).
Here,
x1 = -1 and x2 = 5
y1 = 4 and y2 = 4
<u>The distance formula is given by :</u>
Distance = 
<u>To find the distance between these two points :</u>
The distance formula is used,
Distance = 
⇒ √(6)²
⇒ √36
⇒ 6
Therefore, the between the points is 6 units.