Given a table <span>representing
the probability distribution of the number of times the John Jay wifi
network is slow during a week. We call the random variable x.
![\begin{tabular} {|c|c|c|c|c|c|c|c|} x&0&1&2&3&4&5&6\\[1ex] p(x)&.08&.17& .21& k& .21& k& .13 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7C%7D%0Ax%260%261%262%263%264%265%266%5C%5C%5B1ex%5D%0Ap%28x%29%26.08%26.17%26%20.21%26%20k%26%20.21%26%20k%26%20.13%0A%5Cend%7Btabular%7D)
Part A:
The total value of p(x) = 1.
Thus, </span><span>
.08 + .17 + .21 + k + .21 + k + .13 = 1
0.8 + 2k = 1
2k = 1 - 0.8 = 0.2
k = 0.2 / 2 = 0.1
Therefore,
the value of k is 0.1Part B:
The expected value of x is given by

Therefore,
the expected value of x is 3.01Part C:
</span><span>The expected value of

is given by

Therefore,
the expected value of
is 12.45</span>
Part D:
The variance of x is given by

Therefore,
the variance of x is 3.39.
Part E
<span>The standard deviation of x is given by

Therefore,
the standard deviation of x is 1.84.
Part F:
The variance of ax, where a is a constant is given by

Thus, the variance of 3x is given by

Therefore,
the variance of 3x is 30.51.
Part G:
The probability that the network has no more that 4 slow times in one week is given by

Since, the </span>network slowness is independent from week to week, the <span>probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is given by

Therefore, </span>
the probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is 0.27Part H:
The variance of x^2 is given by


Thus,

Therefore,
the <span>
variance of the random variable
is 141.37</span>
Answer:
25
Step-by-step explanation:
The median of a trapezoid equals one half of the sum of the bases
AC is a median and EB and DF are the bases.
hence AC = 1/2(EB + DF)
We are given that EB = 13 and that AC = 19. And we need to find DF
To do so we plug in what we are given and solve for DF
AC = 1/2(EB + DF)
AC = 19, EB = 13
19 = 1/2(13 + DF)
Now solve for DF
* Multiply both sides by 2*
19 * 2 = 38
1/2(13 + DF) * 2 ( the 1/2 and 2 cancel out and we're left with 13 + DF )
We then have 38 = 13 + DF
* Subtract 13 from both sides *
38 - 13 = 13 - 13 + DF
We get that DF = 25
Answer:
80 km
Step-by-step explanation:
First, find the perimeter of the park.
2(12 + 8) = 2(20) = 40 km
Since Rama travels around the park twice, multiply the perimeter by 2.
40(2) = 80 km