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3 years ago

I will rate you brainliest i need help with these 3 questions

Mathematics

1 answer:

Vlad1618 [11]3 years ago

4 0

Answer:

1) difference of squares, then its factors are (x+4)(x+2)

2) 3 (x+4)/2 (x-4)(x+4)

3/2 (x-4)

3) 3x+11/5x-9

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-t+5=t-19<br> Need an answer and a work pls!

Nadusha1986 [10]

Answer:

t=12

Step-by-step explanation:

Step 1: Subtract t from both sides.

−t+5−t=t−19−t

−2t+5=−19

Step 2: Subtract 5 from both sides.

−2t+5−5=−19−5

−2t=−24

Step 3: Divide both sides by -2.

−2t−2=−24−2

8 0

3 years ago

Five kittens are sharing 6 cups of milk equally. How much milk does each kitten get?

Llana [10]

about 0.83 just solve the equation 5 divided by 6

3 0

3 years ago

Read 2 more answers

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

HELP PLEASE ! ASAP !

Blizzard [7]

Answer: i think 23x

Step-by-step explanation: i might be incorrect

4 0

3 years ago

Read 2 more answers

Aahuti bought a mobile for Rs.16000 and solve to Aarati at a profit of 20%.Aarati sold it to Aakriti with 10%loss. find out the

AveGali [126]

Answer:

1) 19200- for Aarati 2) 17280- for Aakriti

Step-by-step explanation:

1. 16000/100*20=3200- THE MONEY WHICH ARE THE PROFIT,

2. 16000+3200=19200- THE PRICE for Aarati

3. 100-10=90 PERCENTS - the price for AAkriti contains from the price from Aarati

4. 19200*(90/100)= 17280- the price for Aakriti

4 0

2 years ago