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3 years ago

More help :3 please and thnx

Mathematics

1 answer:

Mariana [72]3 years ago

6 0

Answer:

a. there's a lot of options but here are a few: 1 and 5, 5 and 6, 2 and 1, 2 and 6

b. also a lot of options but here are a few: 1 and 6, 5 and 2, 3 and 8, 4 and 7

Step-by-step explanation:

supplementary angles are two angles that add up to 180 degrees, so essentially two angles that, combined, are equal to a straight line.

vertical angles are angles that are opposite each other when two lines cross.

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Julio wanted to borrow $2400 from his parents to buy a scooter. His parents told him they would loan him the money for only 3% s

velikii [3]

$2700 is the answer

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2 years ago

Help:) Factor f(x) = 4x^3+ 23x^2- 70x + 16 into linear factors given that - 8 is a zero of f(x).

Charra [1.4K]

Answer:

(x + 8)(x - 2)(4x - 1)

Step-by-step explanation:

Given that x = - 8 is a zero then (x + 8) is a factor

Divide f(x) by (x + 8) using Synthetic division

- 8 | 4 23 - 70 16

↓ - 32 72 - 16

---------------------------------

4 - 9 2 0

Quotient = 4x² - 9x + 2 = (x - 2)(4x - 1)

Thus

f(x) = (x + 8)(x - 2)(4x - 1)

8 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.

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2 years ago

Which ones correct?

natima [27]

Answer:

Step-by-step explanation:

y = 4 is a horizontal line which would be parallel to the x axis

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2 years ago

The price of a game thats been discounted 18% off its price(x)?

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2 years ago