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4 years ago

Jim has calculated the area of a rectangle to be x^3 + x^2 + x + 1. If the width of the rectangle is x + 1, then its length is

1 answer:

4 0

A = <span>x^3 + x^2 + x + 1
A = x^2(x +1) + (x+1)
A = (X + 1)(x^2 + 1)

W = x +1
so L = x^2 + 1 because A = L*W

answer
</span><span>A) x^2 + 1 </span>

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What is the average price of an item sold considering quantity? 12 notepads at $4.20 25 folders at $0.80 3 calculators at $19.00

Savatey [412]

Answer:

Average price = $3.185

Step-by-step explanation:

Given:

12 notepads = $4.20 each

25 folders = $0.80 each

3 calculators = $19.00 each

Find:

Average price

Computation:

Average price = Total cost / Total number of item

Average price = [(12)(4.20) + (25)(0.80) + (3)(19)] / [12 + 25 + 3]

Average price = [(50.4) + (20) + (57)] / [40]

Average price = [127.4] / [40]

Average price = $3.185

7 0

3 years ago

Substitute t=3 and t=5 to determine if the two expressions are equivalent.

motikmotik

Answer:

because both sides are equal when substituted by 3 & 5

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3 years ago

What is the GCF of 24 and 40

nikitadnepr [17]

Answer:

Step-by-step explanation:

24 = 2* 2 * 2 * 3

40 = 2 * 2 * 2 * 5

GCF = 2*2*2 = 8

6 0

3 years ago

Solve: 6 – x = 4(5 + x) – 4

frez [133]

Answer:

x= -2

Step-by-step explanation:

3 0

3 years ago

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.

8 0

3 years ago