Ray’s weight increased by 11% in the last two years. If he gained 16.5 pounds, what was his weight two years ago?

Solve 3^2x+3=243 for x. A. x = 10 B. x = 2 C. x = 2/5 D. x = 1

telo118 [61]

I think its B but im not 100% sure

3 0

3 years ago

In what time the compound amount of Rs 60000 at 10% rate is rs 79860?

Ksenya-84 [330]

Answer:

P= Rs 60000

A= Rs 79860

T=1 & 1/2 year = 3/2 years

= 3/2 x 2 = 3 half years

R= ?

Applying the formula A= P (1+r/100)^T

79860 = 60000 (1+ r/100)^3

79860/60000 = (1+r/100)^3

1331/1000 = (1+r/100)^3

root(3)(1331/1000) = (1+r/100)

11/10 = 1+r/100

11/10 -1 = r/100

1/10 = r/100

r= 10 %

Step-by-step explanation:

8 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

PLS HELPPP WILL AWARD BRAINLIEST!

AfilCa [17]

Answer: Red is 0.18 And Blue is 0.68

Step-by-step explanation:

Look at sizes and comparisons between the sizes and all together should add to 1

4 0

3 years ago

Read 2 more answers

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2 h, and Car B traveled the dista

irakobra [83]

Answer: Car B traveled 60 mph.

Step-by-step explanation:

A= B-15

A: (B - 15)(2 hr)=D

B: (B)(1.5)=D

Since B and A traveled the same distance you can plug in 1.5B for D.

2B - 30 = 1.5B

-1.5B +30 = -1.5B

——————+30—

2(0.5B) = 2(30)

B = 60 mph

4 0

3 years ago

Read 2 more answers