What percent of 292 is 73

1. At a certain company, the human resources department surveyed employee satisfaction. They collected the following data. (a) H

PilotLPTM [1.2K]

A] From the graph, there is strong correlation between the satisfaction score and salary, this is implied by the closeness of the points on the graph.
b] Given that Hilda use the function y=0.0005x +60 to model the relationship, the score for $65000 will be given as follows;
y=0.0005(65000)+60
y=92.5
The number 0.0005 shows the rate of change of job satisfaction with salary

4 0

3 years ago

A change purse contains an equal number of pennies, nickels, and dimes. The total value of the coins is 208

Aliun [14]

Answer:

x= 14

Step-by-step explanation:

x + 5x +10x = 224 CENTS

solving for x

16x = 224

x = 14, the number of each coin there is

5 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

Clodagh and Eva went school supply shopping together. Clodagh bought 3 packs of pencils and 5 notebooks for a total of $18. Eva

elena55 [62]

Answer:

each notebook costs $2.70

each pack of pencils costs $1.50

Step-by-step explanation:

system of equations:

let p = pack of pencils

let n = notebook

3p + 5n = 18

4p + 4n = 16.8

I used the elimination method by multiplying the first equation by 4 and the second equation by -3

4(3p + 5n = 18) = 12p + 20n = 72

-3(4p + 4n = 16.8) = -12p -12n = -50.4

adding the new equations together you get: 8n = 21.6

n = 21.6/8

n = $2.70

solve for 'p':

3p + 5(2.7) = 18

3p + 13.5 = 18

3p = 4.5

p = $1.50

4 0

3 years ago

Tell whether each scale reduces, enlarges, or preserves the size of the actual object.

liubo4ka [24]

1m is equal to 100cm, so the scale reduces the size of the actual object

5 0

3 years ago

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