Reflection of a figure is called the post image true or false

suppose that the lifetime of a transistor is a gamma random variable x with mean of 24 weeks and standard deviation of 12 weeks.

emmainna [20.7K]

The probability that the transistor will last between 12 and 24 weeks is 0.424

X= lifetime of the transistor in weeks E(X)= 24 weeks

O,= 12 weeks

The anticipated value, variance, and distribution of the random variable X were all provided to us. Finding the parameters alpha and beta is necessary before we can discover the solutions to the difficulties.

X~gamma( $\alpha ,\beta$ )

E(X)= $\alpha \beta$ $\beta$ = $12^{2}/24$ =6 weeks

V(x)= $\alpha \beta ^{2}$ $\alpha$ =24/6= 4

Now we can find the solutions:

The excel formula used to create Figure one is as follows:

=gammadist(X, $\alpha$ , $\beta$ , False)

P( $12\leq X\leq 24$ )

P( $12/6\leq G\leq 24/6$ )

P( $2\leq G\leq 4$ )

P= 0.424

Therefore, probability that the transistor will last between 12 and 24 weeks is 0.424

To learn more about probability click here:

brainly.com/question/11234923

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4 0

1 year ago

Read 2 more answers

Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1

sasho [114]

I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

$\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1$

Where $(x_{1},y_{1})$ is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

$(-3,5)$

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.

The major axis in this case is that of the y-axis. In other words,

$b^2>a^2$

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

$(-3,10)$
& $(-3,0)$

I really hope this helped you! (Partially because I spent a lot of time on this lol)

Sincerely,

~Cam943, Junior Moderator

6 0

3 years ago

Write an equation of a parabola that opens to the left, has a vertex at the origin, and a focus at (–4, 0).

8_murik_8 [283]

Answer:

$y^{2}=-16x$