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3 years ago

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Thomas bikes at a rate of 24 mi in 3 h. Thomas’s friend De’Monte bikes 9.5 mi each hour. De’Monte and Thomas are 105 mi apart an

d are biking towards each other. How long will the boys bike before they meet? How many miles will each boy have biked when they meet?

1 answer:

Murrr4er [49]3 years ago

8 0

6 hours
...............................

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One side of a parallelogram has endpoints (3, 3) and (1, 7).. . What are the endpoints for the side opposite?. . . . (8, 1) and

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The endpoints for the opposite side would be (8, 1) and (6, 5). The correct answer between all the choices given is the last choice. I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.

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Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

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I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

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3 years ago

4. A value of 500 increases by 12% What is the new value?"

Nonamiya [84]

Answer:

increase percent=12%

hom much increase

500 of 12%

500×12/100=60

new value=500+60

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For the following system, if you isolated x in the first equation to use the substitution method, what expression would you subs

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Answer:

$x=2y+6$

Step-by-step explanation:

So we have the system:

$-x+2y=-6\\3x+y=8$

If we isolate the x-variable in the first equation:

$-x+2y=-6$

Subtract 2y from both sides:

$-x=-6-2y$

Divide both sides by -1:

$x=2y+6$

Therefore, we would substitute the above into the second equation:

$3x+y=8\\3(2y+6)+y=8$

The answer is 2y+6

Further notes:

To solve for the system, distribute:

$6y+18+y=8$

Simplify:

$7y+18=8$

Subtract:

$7y=-10$

Divide:

$y=-10/7\approx-1.4286$

Now, substitute this value back into the isolated equation:

$x=2(-10/7)+6\\x=-20/7+42/7\\x=22/7\approx3.1429$

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