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Anit [1.1K]
1 year ago
14

Which point is a solution to the system of inequalities graphed here? y ≤ 2x 2 y ≥ -5x 4

Mathematics
1 answer:
Mandarinka [93]1 year ago
7 0

The point that is a solution to the system of inequalities is (5, 0)

<h3>How to determine the points on the solution?</h3>

The system of inequalities is given as:

y ≤ 2x + 2

y ≥ -5x + 4

Next, we test the given options.


From the list of options, we have

(x, y) = (5, 0)

Substitute (x, y) = (5, 0) in y ≤ 2x + 2 and y ≥ -5x + 4

y ≤ 2x + 2

0 ≤ 2 * 5 + 2

0 ≤ 12 -- this is true

y ≥ -5x + 4

0 ≥ -5 * 5 + 4

0 ≥ -21 -- this is true

Since both results are true, then it means that the point that is a solution to the system of inequalities is (5, 0)

Read more about system of inequalities at:

brainly.com/question/24372553

#SPJ1

<u>Complete question</u>

Which point is a solution to the system of inequalities graphed here? y ≤ 2x + 2

y ≥ -5x + 4

A. (1,6)

B. (-6,0)

C. (0,5)

D. (5,0)

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Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime x (in weeks) has a gamma
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a) P(1 \leq X \leq 40)

In order to find this probability we can use excel with the following code:

=GAMMA.DIST(40;5,8,TRUE)-GAMMA.DIST(1,5,8,TRUE)

And we got:

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b) P(X \geq 40)=1-P(X

In order to find this probability we can use excel with the following code:

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And we got:

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Step-by-step explanation:

Previous concepts

The Gamma distribution "is a continuous, positive-only, unimodal distribution that encodes the time required for \alpha events to occur in a Poisson process with mean arrival time of \beta"

Solution to the problem

Let X the random variable that represent the lifetime for transistors

For this case we have the mean and the variance given. And we have defined the mean and variance like this:

\mu = 40 = \alpha \beta  (1)

\sigma^2 =320= \alpha \beta^2  (2)

From this we can solve \alpha and [/tex]\beta[/tex]

From the condition (1) we can solve for \alpha and we got:

\alpha= \frac{40}{\beta}    (3)

And if we replace condition (3) into (2) we got:

320= \frac{40}{\beta} \beta^2 = 40 \beta

And solving for \beta = 8

And now we can use condition (3) to find \alpha

\alpha=\frac{40}{8}=5

So then we have the parameters for the Gamma distribution. On this case X \sim Gamma (\alpha= 5, \beta=8)

Part a

For this case we want this probability:

P(1 \leq X \leq 40)

In order to find this probability we can use excel with the following code:

=GAMMA.DIST(40;5,8,TRUE)-GAMMA.DIST(1,5,8,TRUE)

And we got:

P(1 \leq X \leq 40)=0.560

Part b

For this case we want this probability:

P(X \geq 40)=1-P(X

In order to find this probability we can use excel with the following code:

=1-GAMMA.DIST(40,5,8,TRUE)

And we got:

P(X \geq 40)=1-P(X

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