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postnew [5]
3 years ago
5

Find the area lying above the x-axis and below the parabolic curve y = 4x -x2 A. 8 B. 16 C. 10 2/3 D. 8 1/3

Mathematics
1 answer:
dimaraw [331]3 years ago
7 0
To find the area between the x-axis and the parabolic curve, take the integral of the area in which the curve is above the x-axis.
function of the graph is
y = 4x - x²
We can tell by the function (specifically -x²) that the parabola will point downward.
To find the domain in which y>0, let's find the roots (0's) of the function:
0 = 4x - x²
0 = x (4 - x) 
x = 0 or x = 4
Between x=0 and x=4, the curve is above the x-axis. To find the area of the graph, let's take the integral on this range:

First, take the antiderivative of 4x - x²:
2x² - (1/3) x³
Then, plug x=4 into the anti-derivative, and subtract the anti-derivative at x=0:
2(4)² - (1/3)(4³) - (0 - 0)
32 - 64/3
96/3 - 64/3 = 32/3 

Closest Answer is C) 10
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5.53 An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day. What
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Answer:

a) 38.4% probability that on a given day this item is requested more than 5 times.

b) 0.67% probability that on a given day this item is not requested at all.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day.

This means that \mu = 5

What is the probability that on a given day this item is requested

(a) more than 5 times?

Either it is requested 5 times or less, or it is requested more than 5 times. The sum of the probabilities of these events is decimal 1. So

P(X \leq 5) + P(X > 5) = 1

We want P(X > 5). So

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067

P(X = 1) = \frac{e^{-5}*(5)^{1}}{(1)!} = 0.0337

P(X = 2) = \frac{e^{-5}*(5)^{2}}{(2)!} = 0.0842

P(X = 3) = \frac{e^{-5}*(5)^{3}}{(3)!} = 0.1404

P(X = 4) = \frac{e^{-5}*(5)^{4}}{(4)!} = 0.1755

P(X = 5) = \frac{e^{-5}*(5)^{5}}{(5)!} = 0.1755

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 + 0.1755 = 0.616

P(X > 5) = 1 - P(X \leq 5) = 1 - 0.616 = 0.384

38.4% probability that on a given day this item is requested more than 5 times.

(b) not at all?

P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067

0.67% probability that on a given day this item is not requested at all.

3 0
3 years ago
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