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Alisiya [41]
3 years ago
11

What are the x- and y- intercepts for the graph of 3x+y=15

Mathematics
1 answer:
allochka39001 [22]3 years ago
6 0
To make the equation easier to look at change it to this: 
y = 15 - 3x 
To find the x intercept say y = 0. 
0 = 15 - 3x 
Subtract 15 from each side. 
-15 = -3x 
Divide each side by -3. 
5 = x <–––– x-intercept
The y-intercept of the equation is 15.
Hope this helps! 
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PIT_PIT [208]
You could say invitability or a foregone conclusion. Invitability just means "the quality of being certian to happen."

Example: "My teacher said it was invitable that she was going to give us a math test tomorrow"
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3 years ago
Is 7/8 greater than or less than 2/5
belka [17]

Answer:

7/8 > 2/5

Step-by-step explanation:

7/8 is greater than 2/5.

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
3. A bag of Jelly beans contains 12 red, 17 white and 14 black jelly beans.
Nataly_w [17]

Answer:

steps below

Step-by-step explanation:

Total = 12+17+14 = 43

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                           ≈ 0.082%

with replacement: 12/43*12/43*12/43*12/43*12/43 = 12⁵ / 43⁵

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2 years ago
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Gala2k [10]

Answer:

63 shelves

Step-by-step explanation:

2986 ÷ 48 = 62. 03

ans = 63

3 0
3 years ago
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