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bagirrra123 [75]
3 years ago
7

What is x / 15 equals 5. what is x?​

Mathematics
2 answers:
MissTica3 years ago
8 0

Answer:

x = 75

Step-by-step explanation:

x/15 = 5

Multiply each side by 15

15*x/15 = 5*15

x = 75

balu736 [363]3 years ago
8 0

Answer:

x = 75

Step-by-step explanation:

Reverse the problem meaning you have to do multiplication

5 times 15 is 75

To check your work divided 75 by 15 and if it equals 5 it is correct

Hope this helped :]

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Based upon a long period of record keeping the following represents the probability distribution of the number of times the John
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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}&#10;{|c|c|c|c|c|c|c|c|}&#10;x&0&1&2&3&4&5&6\\[1ex]&#10;p(x)&.08&.17& .21& k& .21& k& .13&#10;\end{tabular}



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 0.1



Part B:

The expected value of x is given by

E(x)=\Sigma&#10; xp(x) \\  \\ =0(0.08)+1(0.17)+2(0.21)+3(0.1)+4(0.21)+5(0.1)+6(0.13) \\ &#10; \\ =0+0.17+0.42+0.3+0.84+0.5+0.78=3.01

Therefore, the expected value of x is 3.01



Part C:

</span><span>The expected value of x^2 is given by

E(x^2)=\Sigma x^2p(x) &#10;\\  \\ =0^2(0.08)+1^2(0.17)+2^2(0.21)+3^2(0.1)+4^2(0.21)+5^2(0.1)+6^2(0.13) \\  \\ &#10;=0(0.08)+1(0.17)+4(0.21)+9(0.1)+16(0.21)+25(0.1)+36(0.13) \\  \\ =0+0.17+0.84+0.9+3.36+2.5+4.68=12.45

Therefore, the expected value of \bold{x^2} is 12.45


</span>
Part D:

The variance of x is given by

Var(x)=E(x^2)-(E(x))^2 \\  \\ =12.45 - (3.01)^2=12.45-9.06 \\  \\ =3.39

Therefore, the variance of x is 3.39.



Part E

<span>The standard deviation of x is given by

\sqrt{Var(x)} = \sqrt{3.39} =1.84

Therefore, the standard deviation of x is 1.84.



Part F:

The variance of ax, where a is a constant is given by

Var(aX)=a^2Var(X)

Thus, the variance of 3x is given by

Var(3X)=3^2Var(X)=9(3.39)=30.51

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Part G:

The probability that the network has no more that 4 slow times in one week is given by

P(x\leq4)=P(0)+P(1)+P(2)+P(3)+P(4) \\  \\ =0.08+0.17+0.21+0.1+0.21=0.77

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