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earnstyle [38]
3 years ago
8

Emilia bought 9/12 pound of sliced salami at the deli counter.Which of the following decimals did the scale show?

Mathematics
1 answer:
Daniel [21]3 years ago
8 0
To start this problem, what we need to do is try to simplify the fraction into one we might know already. The best way to go about doing this is to see if we can divide the top and bottom by any of the same numbers. 

We can divide 9 by 1, 3, and 9
We can divide 12 by 1, 3, 4, and 12

Looking back over the  numbers, we can see that both 9 and 12 are divisible by 3, so if we divide both of them by 3, we will get a new fraction! This fraction will be 3/4.

Our next step is to see if we know what 3/4 is. I know 1/4 is .25 because the coin called a quarter is 25 cents, so if I had 3 quarters, I would have 75 cents, so I know know that 9/12 = 3/4 = .75!
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4 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
Please answer, need it today! ASAP!
Montano1993 [528]

There is no picture.....

???

5 0
3 years ago
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