Divide both by a number that can go into both terms so it will be 5 .

which is 5(c-3)
Answer:
The probability that a randomly selected call time will be less than 30 seconds is 0.7443.
Step-by-step explanation:
We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.
Let X = caller times at a customer service center
The probability distribution (pdf) of the exponential distribution is given by;

Here,
= exponential parameter
Now, the mean of the exponential distribution is given by;
Mean =
So,
⇒
SO, X ~ Exp(
)
To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;
; x > 0
Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)
P(X < 30) =
= 1 - 0.2557
= 0.7443
The fraction can't be simplify since it already is
Answer:
we can conclude that there is no significant evidence to conclude that the mean score in 2010 differs from the mean score in 2009.
Step-by-step explanation:
H0 : μ = 582
H1 : μ < 582
Test statistic :
T = (xbar - μ) ÷ σ/√n
Xbar = 515 ; n = 20 ; σ = 120
T = (515 - 582) ÷ 120/√20
T = -67 / 26.832815
T = 2.50
Pvalue at t score = 2.50 ; df = 19 is 0.0187
At α = 0.0187
Pvalue > α ; Hence, we fail to reject the Null
Hence, we can conclude that there is no significant evidence to conclude that the mean score in 2010 differs from the mean score in 2009.
Answer:
5
Step-by-step explanation:

<em>G</em><em>CF</em><em> </em><em>=</em><em> </em><em>5</em>