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Kitty [74]
4 years ago
6

Given the following probability distribution, what is the expected value of the random variable X? X P(X) 100 .10 150 .20 200 .3

0 250 .30 300 .10 Sum1.00 Multiple Choice 175 150 200 205
Mathematics
1 answer:
Mekhanik [1.2K]4 years ago
5 0

Answer:

d) 205

Step-by-step explanation:

<u><em>Step(i):-</em></u>

x             :   100      150      200      250      300

p(X=x)    :   0.10     0.20    0.30     0.30      0.10

<u>Step(ii):-</u>

Let 'X' be the discrete random variable

Expected value of the random variable

   E(X) = ∑ x P(X=x)

           =  100 X 0.10 + 150 X 0.20 + 200 X 0.30 +250 X 0.30 + 300 X 0.10

          =   205

<u>Final answer:</u>-

The expected value     E(X) = 205

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Answer:

483 840 ways

Step-by-step explanation:

Let you and your friend be A and B.

With three people always between you two,, we have

Case 1: AxxxBxxxxx

Case 2: xAxxxBxxxx

Case 3: xxAxxxBxxx

Case 4: xxxAxxxBxx

Case 5: xxxxAxxxBx

Case 6: xxxxxAxxxB

So there is 6 ways of arranging you two through the line and for each of the six cases,

If you two remained in your positions, the rest can be arranged in 8! ways

Also for each case, you two can interchange your positions in 2! ways

Therefore,

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3 years ago
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3 years ago
In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
Can you help me with this question please
Aleksandr-060686 [28]

Answer:

The actual length of the lap = 250 meters

Step-by-step explanation:

∵ The scale factor of the map is 1:1000

∵ The length of the lap on the map is 25 cm

∴ \frac{1}{1000}=\frac{25}{x}

Where x is the actual length of a lap

∴ x = (25 × 1000) ÷ 1 = 25000 cm

∵ 1 meter = 100 centimeter

∴ The actual length of the lap = 25000 ÷ 100 = 250 meters

7 0
4 years ago
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Since 7.6 is the cost per package, and x represents the number of packages, the expression 7.6x would represent the cost for the packages. Once we add the base fee, we get the equation:

y = 7.6x + 6.95

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5 0
3 years ago
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