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3 years ago

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Lulu has raised 3/4 of the money she needs for a field trip. 1/2 of that money came from candy sales. What fraction of the total

money needed for the field trip came from candy sales?

2 answers:

iVinArrow [24]3 years ago

7 0

Answer with explanation:

Amount of money raised by Lulu for a field trip is $\frac{3}{4}$

Amount of money that came from Candy Sales is $\frac{1}{2}$ .

The fraction of total money needed for the field trip came from candy sales

$=\frac{1}{2} \text{of} \frac{3}{4}\\\\=\frac{1}{2}*\frac{3}{4}\\\\=\frac{3}{8}$

Papessa [141]3 years ago

3 0

3/8 of the money came from the candy sale

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Answer:

The expected value of X is $E(X)=\frac{2754}{73} \approx 37.73$ and the variance of X is $Var(X)=\frac{226192}{5329} \approx 42.45$

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Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or $\mu_x$ , is

$E(X)=\sum_{x\in D} x\cdot p(x)$

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The expected value of X is

$E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)$

$E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73$

The expected value of Y is

$E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)$

$E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5$

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

$V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2$

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$E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)$

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