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Alekssandra [29.7K]

3 years ago

15

Which is the correct simplified version of the expression -7(2x-3)

1 answer:

satela [25.4K]3 years ago

7 0

A. -14x+21

because -7 times 2x is -14x and -7 times -3 is 21

You might be interested in

Complete the identity. =( sin (alpha +beta))/ (cos alpha cos beta)=?

dem82 [27]

Hi, We know:

Sin(a + b) = 2.Sin(a).Sin(b)

Then,

= 2Sin(a).Cos(b)/Cos(a).Cos(b)

Canceling Cos(b)

= 2Sin(a)/Cos(a)

As Tan(a) = Sin(a)/Cos(a)

Then we will stay:

= 2Tan(a)

Or

= Tan(a) + Tan(a)

There is anything wrong on the alternatives

4 0

3 years ago

A) work out the radius of the mercury, giving your answer as an ordinary number

guapka [62]

Answer:

a. 2,440 km

b. $1.33*10^{27} kg$

Step-by-step explanation:

a. Diameter of Mercury = $4.88*10^3$

Radius of Mercury = ½ of its diameter = $\frac{4.88*10^3}{2}$

Radius of Mercury = $2.44*10^3$

To convert the radius to ordinary number, multiply 2.44 by 1,000.

Radius of Mercury as an ordinary number = $2.44*1,000 = 2,440 km$

b. Mass of Jupiter = $1.898*10^{27} kg$

Mass of Saturn = $5.68*10^{26} kg$

Difference between mass of Jupiter and mass of Saturn = $1.898*10^{27} - 5.68*10^{26}$ (subtract 5.68 from 1.898 = 5.68. we would retain the largest exponent, which is raised to the lower of 27.)

$= 1.33*10^{27}$

7 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

What is your favorite color of the alphabet true or false???

valina [46]

Trueeeeee i guess oop

7 0

3 years ago

balu736 [363]

Answer:

A

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers