Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
The common difference is 12.
It form a right triangle
legnth of ladder is hypotonuse
bottom of ladder from wall distance is one leg
distance up wall is other leg
a^2+b^2=c^2
c=hypotonuse and a and b are legs
4^2+b^2=20^2
16+b^2=400
minus 16 both sides
b^2=384
sqrt both sides
b=8√6 ft
aprox
b=19.59ft from ground
![\qquad \tt \rightarrow \:Domain = [-9, -1]](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Crightarrow%20%5C%3ADomain%20%3D%20%5B-9%2C%20-1%5D)
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Domain = All possible values of x for which f(x) is defined
[ generally the extension of function in x - direction ]
Range = All possible values of f(x)
[ generally the extension of function in y - direction ]
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![\qquad \tt \rightarrow \: range= [ -1,3]](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Crightarrow%20%5C%3A%20range%3D%20%5B%20-1%2C3%5D)
Answered by : ❝ AǫᴜᴀWɪᴢ ❞
