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3 years ago

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Meg is 6 years older than Victor. Meg's age is 2 years less than five times Victor's age. The equations below model the relation

ship between Meg's age (m) and Victor's age (v):
m = v + 6
m = 5v − 2

Which is a possible correct method to find Meg's and Victor's ages?

1 answer:

Verdich [7]3 years ago

6 0

Since no possible correct method is posted, I will suggest a couple.

Method 1: guess and check

Works well for simple problems involving integers like this one.

Victor's age must be zero or greater than one, say one.

Guess v=1, find m=v+6=7, check m=5v-2=5-2=3 no good.

we need to make v bigger

Guess v=2, find m=v+6=2+6=8, check m=5v-2=5*2-2=8 ✔

So v=2, m=8.

Method 2:

Solve the system of two equations.

since the left-hand sides is m in both equations, and since m=m, we just have to equate the right-hand sides to solve for v.

5v-2=v+6

Solve for v

5v-v = 6+2

4v=8

v=2,

so again, v=2, m=v+6=2+6=8.

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Answer:

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Step-by-step explanation:

We are told that;

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3 years ago

The axis of symmetry for the function f(x) = –2x2 + 4x + 1 is the line x = 1. Where is the vertex of the function located?

egoroff_w [7]

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola. The first few steps are as follows. Set the parabola equal to 0 so you can solve for the vertex. Separate the x terms from the constant by moving the constant to the other side of the equals sign. The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out). Let's start there. The first 2 steps result in this polynomial:

$-2x^2+4x=-1$ . Now we factor out the -2:

$-2(x^2-2x)=-1$ . Now we complete the square. This process is to take half the linear term, square it, and add it to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. We add 1 into the set of parenthesis. But we actually added into the parenthesis is +1(-2). The -2 out front is a multiplier and we cannot ignore it. Adding in to both sides looks like this:

$-2(x^2-2x+1)=-1-2$ . Simplifying gives us this:

$-2(x^2-2x+1)=-3$

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex. Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

$-2(x-1)^2+3=y$

From this form,

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you can determine the coordinates of the vertex to be (1, 3)

5 0

3 years ago

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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

nadya68 [22]

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

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adell [148]

Answer:

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Step-by-step explanation:

You know the total price of 22 copies was $365.66. To find the price of one copy, you divide 365.66 by 22 to get $17.53 per copy.

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3 years ago

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Archy [21]

Answer:

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Step-by-step explanation:

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