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IRISSAK [1]
2 years ago
13

What is the price of gas at the gas station in dollars per liter?

Mathematics
1 answer:
dimaraw [331]2 years ago
8 0

Answer:

The price of gas at the gas station is $0.55 per liter.

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Find the constant of variation for the relation and use it to write an equation for the statement. Then solve the equation. If y
dsp73
Y=k/x²; substitute x=1 and y=1/8: 1/8=k. So y=1/(8x²)
When x=5, y=1/(8×25)=1/200.
5 0
3 years ago
Please help me finish these for summer school :)
ValentinkaMS [17]
13.45 will be the answer

since we are finding the hypotenuse, the formula is a^2 + b^2 = c^2

this will then be 10^2 + 9^2 = c^2

181=c^2

c=13.45 :)
5 0
2 years ago
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Find the difference:<br> x+1/x-1 minus x-8/x-1
PolarNik [594]

Answer: 9/x-1

Step-by-step explanation:

Sine you are subtracting,

x-x=o

1- -8 =9 since subtracting a negative is adding a positive.

So the answer is 9/x-1

6 0
3 years ago
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Simplify the product by using FOIL <br><br> (6x-5)(2x-3)
Lapatulllka [165]

12x²-8x-15 should be your answer

3 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
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