Chris had 3/4 of a jug of milk left. He used 2/5 of it to bake a cake how much of the jug of milk did he use in his cake
1 answer:
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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) =
= 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(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
Take 1960 as 0 then find out next year's
- 1970=10
- 1980=20
- 1990=30
- 2000=40
- 2005=45
- 2010=50
Also
Divide each population by one thousand i.e turn to decimal .
This we done to create the plot on graph easily
The graph has been attached
for the below points
- (0,9.706)
- (10,10.652)
- (20,10.798)
- (30,10.847)
- (40,11.353)
- (45,11.478)
- (50,11.576)
