Find the area of the following quadrilateral.

One kilogram of cheese costs £6.60 how much would 200g cost?

Helga [31]

Answer:

0.132

Step-by-step explanation:

if 1kg of chees cost 6.60 100g cost 0.0660 So if 0.0660 multiplied by 2

= 0.132

7 0

2 years ago

HELP!:

algol [13]

Answer:

she can spend 3.00 dollar on he last chocolate

Step-by-step explanation:

14.00-3.00=11.00

11.00-5.00=6.00

6.00-2.00=4.00

4.00-1.00=3.00

8 0

2 years ago

What is the simplified form of k-5+7

ohaa [14]

Answer:

k+2 or k=2

Step-by-step explanation:

combine -5 and 7 to get your answer

4 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

Let f (x) = -3x + 4. Describe the transformations from the graph of f to the graphs of g(x) = f (x) + 3 and h(x) = f (x - 2).

iris [78.8K]

Answer:

Answer:

g(x)=-6/5x+1/2

h(x)=-6/5x-1/2

Step-by-step explanation:

1). g(x)=−f(x) ?

f(x)=6/5x−1/2

g(x)=−(6/5x−1/2)

g(x)=-6/5x+1/2

2). h(x)=f(−x) ?

f(-x)=6/5(-x)−1/2

f(-x)=-6/5x-1/2

h(x)=-6/5x-1/2

Step-by-step explanation:

5 0

1 year ago