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
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Answer: 5
Step-by-step explanation:
Since
, by the corresponding angles theorem,
. This means
by AA.
As corresponding sides of similar triangles are proportional,

However, as distance must be positive, we consider the positive solution, x=5.
Therefore, the answer is <u>5</u>
<span>A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards , what are its dimensions?
I know the basic formula is 2W+2L=300 but i am not sure where to go from there...
-----
Equations:
2W + 2L = 300
L = 2W
----
Substitute for "L" and solve for "W":
</span><span>2W + 2(2W) = 300
6W = 300
W = 50 yds (width)
----
Solve for "L":
L = 2W
L = 100 yds (length)
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Cheers.
</span>
That point of concurrency is called the incenter of the triangle