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2 years ago

How can you put 33 people into different classes

Mathematics

1 answer:

nika2105 [10]2 years ago

8 0

Answer:

Yes

Step-by-step explanation:

This is already tough due to 33 being an odd number of people. And obviously, an amount of people can't be a decimal.

So in order to do this, simply divide 33/3 which is 11. This would mean there would be 3 people per classroom.

3 people in each of the 11 classrooms!

That's it, solved!

Thank you,

Eddie

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What is the slope of the line that passes through the points (2,-5) and (-6,4) ?

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Answer: B. -9/8

Step-by-step explanation:

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3 years ago

What are some ways to decribe the relationships between sets of numbers

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By their increase or decrease

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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

Randy buys gas at a store that charges $3 per gallon. Sarah buys gas at a different store. The table shown here gives the cost w

marin [14]

Where is the table??

5 0

3 years ago

Select the two figures that are similar to the 5 by 10 figure that is shown.

Lapatulllka [165]

Answer:

Figure (i) and (iv)

Step-by-step explanation:

Given:

Optional figure is given in attached file.

We need to find two figures that are similar to the 5 by 10 figure.

All the given figure are $m\times n$ form.

Where m represent the number of rows and n represent the number of columns.

Solution:

Observe that in the given figure 5 by 10, the number of rows is 5 and number of columns is 10, that is, the number of columns is double of that the number of rows.

So we need to find two such figures whose number of columns is double of the number of rows.

From the given figures, figure (i) the number of rows is 2 and number of columns is 4, which is double of number of rows. so it is similar to 5 by 10 figure.

Similarly in figure (iv), the number of rows is 4 and number of columns is 8. so the number of columns is double the number of rows, so it is similar to the figure 5 by 10.

Therefore, the two figures that are similar to 5 by 10 figure are given in attached file such as (i) and (iv).

6 0

3 years ago