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Naddika [18.5K]
2 years ago
9

Solve 18-23 30+ points

Mathematics
1 answer:
raketka [301]2 years ago
8 0

Answer:

-5

Step-by-step explanation:

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There are 65 people who want to go on a tea cups ride at an amusement park. There are 10 tea cups that each seat 4 people. How m
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25 people will wait for the next ride
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3 years ago
In a sale the original prices are reduced by 20%.
Anettt [7]

Answer:

17260

Step-by-step explanation:

multiply 14400 by 20%

After that add 14400 and 2880together

and you got the answer

6 0
3 years ago
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Jackrabbits are capable of reaching speeds up to 40 miles per hour. How fast is this in feet per second? (Round to the nearest w
Contact [7]
There are 5280 feet in a mile
There are 3600 seconds in an hour

to get 40 miles per hour
we multiply 40 by 5280 and divide by 3600

40 * 5280 / 3600

this gets us 58.666666...
but since they want us to round to the nearest whole number, our answer would be 59 feet per second
5 0
3 years ago
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
The slope of the line is -3. Use the coordinates of the labeled point to find the point-slope equation of the line.
nekit [7.7K]
m =  \dfrac{Y_2 - Y_1}{X_2-X_1}

-3 =  \dfrac{y + 7}{x - 5}

y + 7= -3(x - 5)

Answer A

8 0
3 years ago
Read 2 more answers
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