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Scorpion4ik [409]
2 years ago
5

What I said 4 3/6 - 2 5/6

Mathematics
2 answers:
AlexFokin [52]2 years ago
7 0

Step-by-step explanation: To subtract mixed numbers, first subtract the fractions.

Notice here however that we have 3/6 - 5/6

which will give us a negative fraction.

Since this will cause us a lot of trouble, instead,

let's rewrite the first mixed number.

We can do this by thinking of 4 and 3/6 as 3 + 1 and 3/6

or 3 + 9/6 by changing 1 and 3/6 to an improper fraction.

So 4 and 3/6 can be written as 3 and 9/6.

So we have 3 and 9/6 - 2 and 5/6.

Now subtract the fractions.

9/6 - 5/6 is 4/6.

So we have 1 and 4/6 or 1 and 2/3.

As an improper fraction, the answer is 5/3.

hoa [83]2 years ago
3 0

Answer:

1  2/3

Step-by-step explanation:

The denominators are the same, so we have to borrow from the 4

Borrow  1 in the form 6/6

3 + 6/6+3/6  - 2 5/6

3 9/6   - 2 5/6

1 4/6

Simplify the fraction by dividing the top and bottom by 2

1  2/3

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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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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

\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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