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Oduvanchick [21]
3 years ago
6

9x^3 + 45x^2 - 4x -20 ; x+5 synthetic division

Mathematics
1 answer:
almond37 [142]3 years ago
3 0
                    9x² + 0x - 4
x + 5|9x³ + 45x² - 4x - 20
         <u>9x³ + 45x²               </u>
                           -4x - 20
                           <u>-4x - 20</u>
                                     0
The answer is equal to 9x² - 4.
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The reliability of a piece of equipment is frequently defined to be the probability, P, that the equipment performs its intended
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Answer:

a. y=1

b. Mean= 0.5; variance=0.5

c. In the first instance, the probability that p > 0.95 would mean that we are looking for the area of the rectangle with a height of 1 between 0.95 and 1, which implies it has length 0.05.

Therefore, the area is length × height, which is 0.05 ×1 = 0.05.

Second case, the probability that p is less than 0.95 would then be one minus the probability that p is greater than 0.95, so 1 - 0.05 = 0.95

D. a horizontal line at 20 between 0.90 and 0.95, and will be zero outside of this range.

Step-by-step explanation:

With the use of probability distribution, the probability that the equipment performs has a uniform distribution with minimum 0 and maximum 1.

a) The graph of the probability distribution will be 0 outside of the range of 0 to 1, so therefore y = 0. Inside the interval from 0 to 1, it will be constant (which is a horizontal line) with height 1÷(1-0) = 1, therefore y = 1.

b) The mean of the uniform is (maximum+minimum)/2.

Therefore, (1+0)/2 = 1/2 or 0.5.

The variance of the uniform is

(maximum-minimum)^2/12,

so (1-0)^2/12 = 1/12.

c) Since the probability distribution is rectangular, you can find probabilities by recalling the area of a rectangle which is: area = length × breadth.

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Therefore, the area is length × breadth, which is 0.05 ×1 = 0.05.

In the second case, the probability that p is less than 0.95 would then be one minus the probability that p is greater than 0.95, so 1 - 0.05 = 0.95.

d) If it is known that p is between 0.90 and 0.95, without the value, then we would assume that p has a uniform distribution between 0.90 and 0.95 since p originally had a uniform distribution.

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f(p) =

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0, otherwise.

In a nutshell, this function will have a horizontal line at 20 between 0.90 and 0.95, and will be zero outside of this range.

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Answer:

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