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3 years ago

Find the HCF and LCM of 20 and 38

Mathematics

2 answers:

Anna007 [38]3 years ago

5 0

I want to help but i remembered im too dumb to know the answer im sorry :((

Vika [28.1K]3 years ago

4 0

Answer:

LCM = 380

HCF = 2

Step-by-step explanation:

Least common multiple (LCM) of 20 and 38 is 380

Greatest common factor (GCF) of 20 and 38 is 2

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Solve the inequality for x. Show each step of the solution.10x > 2(8x – 3) – 18

Luden [163]

Answer: x<4

Given the inequality:

$10x>2(8x-3)-18$

We want to solve the inequality for x.

First, distribute the bracket on the right side of the inequality.

$\begin{gathered} 10x>2(8x)-2(3)-18 \\ 10x>16x-6-18 \\ 10x>16x-24 \end{gathered}$

Next, subtract 10x from both sides of the inequality.

$\begin{gathered} 10x-10x>16x-10x-24 \\ 0>6x-24 \end{gathered}$

Add 24 to both sides of the inequality.

$\begin{gathered} 0+24>6x-24+24 \\ 24>6x \end{gathered}$

Divide both sides of the inequality by 6.

$\begin{gathered} \frac{24}{6}>\frac{6x}{6} \\ 4>x \\ \implies x$

The solution to the inequality is x<4.

6 0

1 year ago

What is (-10,8) in point- slope form?

nata0808 [166]

There can't be a point-slope form until you have TWO points.
A single point doesn't have any other form.
There are an infinite number of lines that all go through a single point.
Every one of them has a different slope, and a different point-slope form.

7 0

2 years ago

Megan makes $480 a week and her boss wanted to give her

wel

Answer:

$537.60

Step-by-step explanation:

12% increase is the same as multiplying by 1.12, 480*1.12=537.60

5 0

3 years ago

Solve for x and then find the measure of angle B <br> Angle B=

Savatey [412]

8x + 6=4x + 38
4x = 32
x = 8

<B = 4x + 38 = 4(8) + 38 = 70

5 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