What percent of the bill should you offer as a tip to a salon artist?

What is p ÷ 6 = 9 ? I'm not really understanding it

Shalnov [3]

Let's figure this out. Step by step:

P / 6 = 9

P = 6(9)

P = 54 is your answer.

I hope this helps :)

8 0

3 years ago

HURRY PLEASE 6TH GRADE MATH HERE. A football field is 100 yards in length. How many inches long is a football field? 300 inches

Rufina [12.5K]

Answer:

3,600

Step-by-step explanation:

1 yard= 36 inches

36 inches*100= 3,600

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Brainliest? I am trying to get to a new rank so I can help more people and I would totally appreciate it if you did!

Merry Christmas!

3 0

3 years ago

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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

Someone please help for brainiest and thanks.. No fake answers or you get reported

bagirrra123 [75]

Hello from MrBillDoesMath!

Answer:

12/5

Discussion:

Note that 10 is a common denominator of both denominators:

9/5 = (9*2)/(5*2) = 18/10

- ( -6/10) = + (6/10)

So the original problem is equivalent to

18/10 + 6/10 =

(18 +6)/10 =

24/10 =

(2*12)/ (2*5) => cancelling the common factor "2"

12/5 =

2 and 2 fifths =

22/5 (though this looks like (22)/5!)

Thank you,

MrB

6 0

3 years ago

Elle went to pet smart and bought 4 gold fish and three turtles for $28. later the day, warren went to Petsmart and bought six g

AlekseyPX

It would cost 40$ because you are doing 10 times 4

8 0

2 years ago

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