All categories

2 years ago

Consider these three numbers expressed in scientific notation 2.4 x10^4 , 6.3 x10^5 , and 9.6 x10^7

Mathematics

1 answer:

____ [38]2 years ago

8 0

I considered them, but if you wanted them evaluated, here you go
24,000
630,000
96,000,000

You might be interested in

Please help me answer this.

TiliK225 [7]

8x2+6x-2

I think! because 8,6, and 2 are all divisible by 2

and 3. Factor the quadratic

2(4x-1)(x+1)

6 0

2 years ago

You buy rice at $0.71 a pound. One batch of fried rice requires 3 pounds rice. How much does the rice for one batch cost? Round

Phoenix [80]

Answer:

To make one batch of fried rice you'd need to pay $2.13

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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

Evaluate: 5-3[2+(-5)]2-8

trasher [3.6K]

Answer:

Step-by-step explanation:

5 - 3[ 2 + (-5)] 2 - 8

Subtract 5 from 2 to get −3.

5 - 3 (-3) × 2 - 8

Multiply 3 and −3 to get −9.

5 - (-9 × 2) - 8

Multiply −9 and 2 to get −18.

5 - (-18) - 8

The opposite of −18 is 18.

5 + 18 - 8

Add 5 and 18 to get 23.

23 - 8

Subtract 8 from 23 to get 15.

Hope it helps and have a great day! =D

~sunshine~

7 0

2 years ago

Read 2 more answers

Penelope's MP3 player has 20 megabytes of memory. She has already downloaded 11 megabytesShe will continue to download m more me

8_murik_8 [283]

Answer:

9 MB

Step-by-step explanation:

20 - 11 = 9

7 0

2 years ago