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3 years ago

15

A meuseum added 18 painting to its collection in may. This is 20% of paintings the meuseum has in April. How many paintings did

the meuseum have in April ?

1 answer:

Alexus [3.1K]3 years ago

5 0

18 / x = 20 / 100

20x = 1800

x = 90

The museum had 90 paintings in April.

Hope this helps!! :)

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MA_775_DIABLO [31]

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Feliz [49]

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3 years ago

Read 2 more answers

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

elena-14-01-66 [18.8K]

Answer:

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:

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

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

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

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

And we got:

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

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 $\alpha$

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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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frutty [35]

Answer:

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