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2 years ago

The Bilbo Company plans to manufacture decals for children's lunchboxes. Each decal

Mathematics

1 answer:

Gwar [14]2 years ago

4 0

Answer: 1 -2ax²+2x-ax

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula:

Area of a rectangle: length x width

Replacing with the values given:

A = (2x+1) (1-ax)

A = [2x(1)] + [2x(-ax)] +[ 1(1)]+ [1 (-ax)]

A = 2x- 2ax² +1- ax

A = 1 -2ax²+2x-ax

In conclusion, the correct option is 1 -2ax²+2x-ax

Feel free to ask for more if needed or if you did not understand something.

You might be interested in

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

If you know the rest of the answers pls help me

netineya [11]

Answer:

.45 times 6= 2.7

29 times 8 =232

.62 times 19 = 11.78

8.7 times 12=104.4

11.9 times 4 = 47.6

Step-by-step explanation:

3 0

1 year ago

Idk how to solve this find value of x and y

uysha [10]

Answer:

Try out each different answers and see which one would work.

Step-by-step explanation:

Go to A, and then B and then C and then D to find the correct one. Plug in the answers to find out which one is correct.

7 0

2 years ago

Sue wrote a double fact. It has a sum less than 10 and greater than 4. The addends are each less than 5. What fact might she hav

zloy xaker [14]

It could be:
3+4=7
2+4=6
1+4=5

8 0

2 years ago

Gratest common factor of the numbers 35 63 84

Ierofanga [76]

$35,63,84 \\ 35: 1,5,7,35\\ 63: 1,3,7,9,21,63 \\ 84: 1,2,3,4,6,7,12,14,21,28,42,84\\GCF: 7$

$Good \\ Luck \\ On\\ Your \\ Assignment!$

5 0

2 years ago