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3 years ago

If (x2 – 4) ÷ (x + 2) = x – 2, which polynomial should fill in the blank below? (x + 2) ______ = x2 – 4

Mathematics

2 answers:

Oksanka [162]3 years ago

7 0

(x + 2) __x – 2__ = x2 - 4
Answer: x - 2

Talja [164]3 years ago

3 0

We can use the given to find the answer.

(x+2) times x = x^2 and 2 times -2 = -4

Therefore, (x+2)(x-2) = x^2-4

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3 years ago

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larisa [96]

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3 years ago

Find the derivative of the function. y = sin−1(6x + 1)

liraira [26]

If you don't know the derivative of the inverse of sine, you can use implicit differentiation. Apply sine to both sides:

$y=\sin^{-1}(6x+1)\implies\sin y=6x+1$

(true for <em>y</em> between -π/2 and π/2)

Now take the derivative of both sides and solve for it:

$\cos y\dfrac{\mathrm dy}{\mathrm dx}=6$

$\dfrac{\mathrm dy}{\mathrm dx}=6\sec y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\cos\left(\sin^{-1}(6x+1)\right)}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\sqrt{1-(6x+1)^2}}$

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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.

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3 years ago

A 4-kilogram bag of birdseed costs $12.69. What is the unit price, rounded to the

igomit [66]

Answer:

$3.17

Step-by-step explanation:

Take the $12.69 and divide it by the 4 kilograms and you get 3.17. (This is rounded)

3 0

3 years ago