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2 years ago

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Claire is a salesperson who sells computers at an electronics store. She makes a base pay of $80 each day and then is paid a $20

commission for every computer sale she makes. Write an equation for P,P, in terms of x,x, representing Claire's total pay on a day on which she sells xx computers.

1 answer:

Bezzdna [24]2 years ago

5 0

The answer is P= 20x+80

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What is the value of x to the power of 6 + y to the power of 2 when x=1 and y=11

siniylev [52]

Answer: 122

Step-by-step explanation: Substitute the values of <em>x</em> and <em>y</em> into the expression.

$x^6+y^2 \\ \text{ is like } \\ ( )^6 + ( )^2$

"Fill in the blanks" with the numbers given for <em>x</em> and <em>y</em>.

$(1)^6+(11)^2 = 1+121=122$

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3 years ago

Which object has the mass of about one kilogram?

Step2247 [10]

Answer:

A book about a mass of 1 kilogram, but that isn't the only answer.

Explanation:

There's many objects with a mass of one kilogram like one liter of water, a pineapple, bag of sugar etc.

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3 years ago

What is the median number of 4,8,10,5,9

Nady [450]

The answer is 10. because median means middle so 10 is the middle #.

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3 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.

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3 years ago

Evelyn has $524.96 in her checking account. She must maintain a $500 balance to avoid a fee. She wrote a check for $32.50 today.

postnew [5]

Answer:

You answer would be $7.54

Step-by-step explanation:

You need to subtract 32.50 from 524.96. That would be 492.46. Then subtract 492.46 from 500. Hope that helps!!

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3 years ago