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3 years ago

15

After selecting 180 of its employees to from a new division, a company discovers that engineers outnumber analysts by a 3:1 rati

o in the division. Oly analysts and engineers work in the new division. How many more analysts must the company add to the division if it wants to have a 1:1 ratio of engineers to analysts at the new division?

1 answer:

nikdorinn [45]3 years ago

3 0

Answer:

I think its 90

Step-by-step explanation:

3x plus 1x is 4x divide 180 but 4 and get 45 so that meab there are 135 and 45

you need 90 more analysts to have a 1 to 1 ratio

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9. Kevin Durant scored 33 points in a recent game. Here are the outcomes of each

lyudmila [28]

Answer:

C

Step-by-step explanation:

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3 years ago

What is $2.50 times 7?

Sliva [168]

2.5*7 = 17.5. If it's easier for you, multiply 2.5 by 5 then add 2.5*2.

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4 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.

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3 years ago

An unfair coin with Pr[H] = 0.75 is flipped. If the flip results in a head, a student is selected at random from a class of si

Agata [3.3K]

Answer:

72.73% probability of selecting a girl, given the flip resulted inheads

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Coin resulted in heads

Student is selected at random from a class of six boys and sixteen girls.

Desired outcomes:

16 girls, so $D = 16$

Total outcomes:

16+6 = 22, that is, 22 students, so $T = 22$

Probability:

$P = \frac{16}{22} = 0.7273$

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3 years ago

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sammy [17]

Answer: 55 minutes

Step-by-step explanation:

First, you would have to subtract 5 from 225 because there are already 5 monsters there.

225-5= 220

Next, you would have to divide 220 by 4 to find the amount of minutes.

220/4= 55

6 0

3 years ago