Please help me solve thisss

Determine the equation for the given line in

Tema [17]

Answer:

Y= -3/5 - 1

Step-by-step explanation:

if you start on the y intercept an go down once (-1)

then go down three times (-3)

positive 5 so go right 5 times it lands on the point :)

im not a teacher and i took this last year so sorry if my explanation isn't the best

6 0

2 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

2 years ago

Gianna is making up bags of fruit for her soccer team. Each bag contains 2 strawberries, 4 blueberries, and 3 raspberries. If sh

NeTakaya

Answer:

Gianna uses 24 raspberries

Step-by-step explanation:

there are 2 strawberries in each bag, if Gianna uses 16 then she must have 8 bags.

So 8 times 3 is 24 (The 3 is how many raspberries in a bag)

So Gianna uses 24 raspberries

Hope this helps!

7 0

2 years ago

The temperature at 6:00 P.M. was 68°F. This was 5°F lower than the temperature at 3:00 P.M. The temperature at 1:00 P.M. on that

bixtya [17]

76 degrees Fahrenheit

6 0

2 years ago

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Samuel consumed 2161 calories of food on Monday, 2378 calories on Tuesday, and 1841 calories on Wednesday. In order for Samue

german

Number of calories consumed on Monday = 2161
Number of Calories consumed on Tuesday = 2378
Number of Calories consumed on Wednesday = 1841
Daily average intake is 2200
Let calories intake on Thursday be X

Now ,

Average = sum of total calories/ number of calories

=> 2200 = 2161+2378+1841+X/4

=> 2200 x 4 = 6377 + X

=> 8800 = 6377 + X

=> X = 8800-6377

=> X = 2423

So Samuel must consume 2423 calories on Thursday

3 0

2 years ago