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3 years ago

7

Jesse Stroh the coin in her bank she made 7 stacks of 6 diamonds and 8 stacks of 5 nickels she then found one diamond and 1 nick

el how many dimes and nickels does Jesse have in all

1 answer:

fiasKO [112]3 years ago

7 0

So there are 8 stacks of 5 nickels which means that the equation would be simply 8 * 5 which equals 40 nickels. Then, 7 stacks of 6 dies would equal 42. Then she found one more dime and one more nickel which means she has 41 nickels and 43 dimes. In all, she has 84 dimes and nickels.

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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.

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3 years ago

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tekilochka [14]

Okay! You’re a girl right? Lol

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3 years ago

Please help quickly! 20 points

scZoUnD [109]

Answer:

y+3 = 4(x+3)

Step-by-step explanation:

its this answer because to find point slope form you have to do y1 - y2 and in the coordinate provided, -3 is y so y- (-3) or y+3 hope this helped ^u^

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3 years ago

A thermometer is taken from a room where the temperature is 20°C to the outdoors, where the temperature is 3°C. After one minute

IrinaVladis [17]

Answer:

A. T=7.12°C; B. t=3.32min

Step-by-step explanation:

In order to find the answer we can use the Newton's law of cooling:

$y(t)=y0e^{kt}$ where:

y0=initial temperature - room temperature

k=coefficients

t=time

Because the outdoors temperature is 3°C, the initial temperature is 20°C, and because the thermometer reads 11°C after 1min we have:

$11=(20-3)e^{k*1}$

$ln(11/17)=k$

$-0.4353=k$

A. Temperature after an extra minute.

$y(2)=17e^{-0.4353*2}$

$y(2)=7.12$ so the answer is 7.12°C

B. Time in order to read 4°C

$4=17e^{-0.4353*t}$

$ln(4/17)/(-0.4353)=t}$

$3.32min=t$

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3 years ago

Which box plot represents the data? 43, 48, 52, 61, 74, 81, 40, 42, 56, 77, 81, 43?

frutty [35]

The box plot for the set of data
<span>43, 48, 52, 61, 74, 81, 40, 42, 56, 77, 81, 43
</span>will be found by rewriting the data in ascending data, then plotting it. This will give us the box plot in the attached figure.

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3 years ago

Read 2 more answers