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3 years ago

What is the pattern in this problem???

Mathematics

1 answer:

Ne4ueva [31]3 years ago

5 0

A number is divisible by 4 when the last two digits is divisible by 4.

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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

The two cones are congruent. Determine the unknown measures of the cones. A = units B = units C = units D = units

VikaD [51]

Answer:

3.1

4.2

5.2

Step-by-step explanation: Just copy the other cone and find half of the diameter to find the radius the cones are congruent.

6 0

3 years ago

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Please help!! I literally have no idea what to do here.

nika2105 [10]

I believe it's it's d
Sorry if I'm wrong I'm mr lemonade

3 0

2 years ago

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Oliga [24]

Answer:

Part A)

The height of the water level in the rectangular vessel is 2 centimeters.

Part B)

4000 cubic centimeters or 4 liters of water.

Step-by-step explanation:

We are given a cubical vessel that has side lengths of 10cm. The vessel is completely filled with water.

Therefore, the total volume of water in the cubical vessel is:

$V_{C}=(10)^3=1000\text{ cm}^3$

This volume is poured into a rectangular vessel that has a length of 25cm, breadth of 20cm, and a height of 10cm.

Therefore, if the water level is h centimeters, then the volume of the rectangular vessel is:

$V_R=h(25)(20)=500h\text{ cm}^3$

Since the cubical vessel has 1000 cubic centimeters of water, this means that when we pour the water from the cubical vessel into the rectangular vessel, the volume of the rectangular vessel will also be 1000 cubic centimeters. Hence:

$500h=1000$

Therefore:

$h=2$

So, the height of the water level in the rectangular vessel is 2 centimeters.

To find how how much more water is needed to completely fill the rectangular vessel, we can find the maximum volume of the rectangular vessel and then subtract the volume already in there (1000 cubic centimeters) from the maximum volume.

The maximum value of the rectangular vessel is given by :

$A_{R_M}=20(25)(10)=5000 \text{ cm}^3$

Since we already have 1000 cubic centimeters of water in the vessel, this means that in order to fill the rectangular vessel, we will need an additional:

$(5000-1000)\text{ cm}^3=4000\text{ cm}^3$

Sincer 1000 cubic centimeters is 1 liter, this means that we will need four more liters of water in order to fill the rectangular vessel.

3 0

3 years ago

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In which quadrant is (4,-5) located

Vinvika [58]

4 in (4,-5) implies positive x-axis.
-5 in (4,-5) implies negative y-axis.

Since x is positive and y is negative in fourth quadrant, so the answer is

QUADRANT IV (or 4th quadrant.

3 0

3 years ago

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