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3 years ago

The sum of four times a number and 18 is the same as the difference of fourteen and twice a number

Mathematics

1 answer:

Citrus2011 [14]3 years ago

4 0

$4x+18=14-2x \\ 4x=-4-2x \\ 6x=-4 \\ x= \frac{2}{3}$

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Answer! The amount of time it takes to complete a reading assignment might be a function of which of the folowing ?

sladkih [1.3K]

Amount is a function of what means
the amount is dependent on what variable

well, the amount of time to read would be depending on the number of pages or your reading speed

looking at the options, A is clearly the answer

5 0

3 years ago

Read 2 more answers

There are 6 boys and 9 girls in the church's youth group. What type of ratio is boys to girls

mina [271]

I hope this helps you

6/9

2.3/3.3

2/3

2:3

3 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

Quiero saber como se resuelve este problema ? 6+2-a3

Andreyy89

Answer: 8-a^3
You just simply the expression.

6 0

3 years ago

Read 2 more answers

One morning, John drive 5 hours before stopping to eat. After lunch, he. increased his speed by 10 mph. If he completed a 500-mi

Marrrta [24]

So he drove 5hrs, before lunch, now, he was going at speed, say "r", after he ate, he sped up by 10mph, so, his rate is whatever "r" is, plus 10, or " r + 10 ", after lunch

we know the whole trip took 10hrs, so, is 5hrs before lunch and 5hrs after lunch

we also know the whole trip was 500miles, so if he drove, say "d" miles before lunch, after lunch he drove the slack after lunch, or " 200 - d "

recall, your d = rt, distance = rate * time

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before lunch}&d&r&5\\
\textit{after lunch}&500-d&r+10&5
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=5r\\
500-d=(r+10)5\\
----------\\
500-\boxed{5r}=(r+10)5
\end{cases}
\\\\\\
\cfrac{500-5r}{5}=r+10\implies 100-r=r+10$

and I'm sure you know what that is

6 0

3 years ago