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Andrej [43]
4 years ago
15

John is interested in purchasing a multi-office building containing five offices. The current owner provides the following proba

bility distribution indicating the probability that the given number of offices will be leased each year. Number of Lease Offices 0 1 2 3 4 5 Probability 5/18 1/4 1/9 1/18 2/9 1/12 If each yearly lease is $12,000, how much could John expect to collect in yearly leases for the whole building in a given year?(in dollars)
a) E(X) = $23,353.33
b) E(X) = $23,333.33
c) E(X) = $23,273.33
d) E(X) = $23,263.33
e) E(X) = $23,423.33
f) None of the above.
Mathematics
1 answer:
Elenna [48]4 years ago
3 0

Answer:

Option B.

Step-by-step explanation:

The given table is:

Number of Lease Offices :  0           1         2       3        4         5

Probability                         : 5/18     1/4     1/9    1/18     2/9      1/12

The expected probability is

Expected probability = \sum_{i=0}^5 x_{i}p(x_i)

Expected probability = 0p(0)+1P(1)+2P(2)+3P(3)+4P(4)+5P(5)

Expected probability = 0\cdot (\frac{5}{18})+1\cdot (\frac{1}{4})+2\cdot (\frac{1}{9})+3\cdot (\frac{1}{18})+4\cdot (\frac{2}{9})+5\cdot (\frac{1}{12})=\frac{35}{18}

It is given that the yearly lease = $12,000.

The yearly leases for the whole building in a given year is

Yearly leases = \frac{35}{18}\times 12000=23333.3333333\approx 23333.33

Therefore, the correct option is B.

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Answer:

The solutions x^4-5x^2-36=0 are x=3,\:x=-3,\:x=2i,\:x=-2i and the x-intercepts of y=x^4-5x^2-36 are x=3,\:x=-3

Step-by-step explanation:

Finding the solutions to x^4-5x^2-36 means finding the roots, a root is where the function is equal to zero.

The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero.

To find the roots you need to:

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\mathrm{Break\:the\:expression\:into\:groups}\\u^2-5u-36=\left(u^2+4u\right)+\left(-9u-36\right)

\mathrm{Factor\:out\:}u\mathrm{\:from\:}u^2+4u=u\left(u+4\right)

\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9u-36=-9\left(u+4\right)

u^2-5u-36=u\left(u+4\right)-9\left(u+4\right)

\mathrm{Factor\:out\:common\:term\:}u+4\\\left(u+4\right)\left(u-9\right)

u^2-5u-36=\left(u+4\right)\left(u-9\right)=0

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions to the quadratic equation are:

\:u=-4,\:u=9

Substitute back u=x^2, solve for x

x^4-5x^2-36=(u-9)(u+4)=(x^2-9)(x^2+4)

Apply the difference of squares formula

x^4-5x^2-36=(x^2-9)(x^2+4)=(x-3)(x+3)(x^2+4)

(x-3)(x+3)(x^2+4)=0

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions are:

\:x=3,\:x=-3,\:x=2i,\:x=-2i

Because two of the solutions are complex roots the only x-intercepts are x=3,\:x=-3

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Step-by-step explanation:

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