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1 year ago

1) What is the area of the rectangle? 4 yd 7yd

Mathematics

2 answers:

Brums [2.3K]1 year ago

8 0

28 hope this helps :)

guapka [62]1 year ago

6 0

Answer:

Area = 28

Step-by-step explanation:

Area = L x W

L = 7

W = 4

Area=7 x 4

Area = 28

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Jessie is 6 years older than Todd, and Todd is 2 years younger than Ryan.

Ilya [14]

Your answer would be d

3 0

2 years ago

Gina's number is 2 more than sara's. the sum of their numbers is 68. what are their numbers?

ikadub [295]

In order to solve this problem, we transform the statements into algebraic expressions. First, we assign the variables.

Let:

x = Gina’s number

y = Sara’s number

For the first equation, we show that Gina’s number is greater than Sara’s number by 2. For the second equation, we show that the sum of both numbers is 68.

(1) x – y = 2

(2) x + y = 68

We add the two expressions, which result in the expression: 2x = 70. Then we divide 70 by 2 to get the value of x. We then have x = 35. Using the second equation, we solve for y = 68-35. This gives y = 33. To summarize, Gina’s number is 35 while Sara’s number is 33.

8 0

2 years ago

Read 2 more answers

Solve the equation for x in simplest form

spayn [35]

Answer:

A x=1/4

Step-by-step explanation:

2/3x = 1/6

Divide by 2/3

so 1/6 ÷ 2/3

= 1/6 • 3/2

= 3/12

= 1/4

4 0

3 years ago

The sum of the three consecutive integers is -27. What is the product of the smallest and largest of the three integers?

Aleksandr [31]

Answer:

B. 80

Step-by-step explanation:

3 consecutive integers: x-1 , x , x+1

(x-1) + x + (x+1) = -27

3x = -27

x = -9

smallest: -9-1 : -10

largest: -9+1 : -8

product of the smallest and largest of the three integers: (-10)*(-8)=80

4 0

2 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

2 years ago