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Rudiy27
3 years ago
8

WILL GIVE BRAINLIEST! PLS HELP!!!

Mathematics
1 answer:
Dmitrij [34]3 years ago
4 0

Answer:

140m

Step-by-step explanation:

multiply 50 and 4.2 which give you 210 then divide by 1.5 and your answer is 140m

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To solve this equation by elimination, what you would do is multiply one of the equations by -1, or distribute -1 to each term in the equation, any of the 2 equations. Then align the equations and add them together.

-(X + 3y = 3)
-X - 3y = -3

-X - 3y = -3
X + 6y = 3
__________
3y = 0
y = 0/3 = 0.

Now we can solve for x, by simply plugging the value of y into any of the 2 equations.

X + 6y = 3
X + 6(0) = 3
X + 0 = 3
X = 3.

The solution to your system of equations would be (3,0).

Check this by plugging in the point to the other equation and see if it is true.

X + 3y = 3
(3) + 3(0) = 3
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Walter's family plays a quick game together every night after dinner. Walter draws names out of a hat to determine which kind of
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Step-by-step explanation:

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3 years ago
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Jake lives in San Mateo. He bought an item for $787.31 and when the clerk calculated the sales tax it came out to exactly
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Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the
Likurg_2 [28]

Answer:

(a)  moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, P(X)=\frac{1}{6}

(a) Moment generating function can be written as M_{x}(t).

M_x(t)=\sum_{x=1}^{6} P(X=x)

M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}

M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) Now, find E(X) \text { and } E\((X^{2}) using moment generating function

M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)

M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)  

\Rightarrow E(X)=\frac{21}{6}

M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)

M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)

\Rightarrow E\left(X^{2}\right)=\frac{91}{6}  

Hence, (a) moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right).

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

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3 years ago
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Answer:

$1,599

Step-by-step explanation:

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