All categories

3 years ago

When two lines crosses, it is found that the angle opposite each other are thesame sizes. They are known as what

Mathematics

1 answer:

Andreyy893 years ago

3 0

Answer:

vertical angles

Step-by-step explanation:

Opposite angles that are congruent are known as vertical angles.

You might be interested in

Lol again another math solver help pls! LOL

gladu [14]

Answer:

b = 4

Step-by-step explanation:

5-1=5b-4b

4=b

4 0

3 years ago

Read 2 more answers

Change 2/7% to a ratio

andre [41]

(2/7)% =

(2 / 7) / 100 = 

2 / 700 

2 : 698 

1 : 349

7 0

3 years ago

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75.percent f

irinina [24]

Answer:

33%

Step-by-step explanation:

Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.

Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.

So we can now equate the parts of the ryegrass in the mixture as:

0.4X + 0.25(100-X) = 30

<=> 0.4X + 25 - 0.25X = 30

<=> 0.15X = 5

<=> X = 5/0.15 = 500/15 = 100/3

So the weight of mixture X as a percentage of the weight of the mixture

= (weight of X/weight of mixture) * 100%

= (100/3)/100 * 100%

= 33%

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

Sophie can peel all the potatoes by herself in 45 minutes, while it would take Simon 30 minutes to do the job working alone. If

Degger [83]

Answer:

18 minutes

Step-by-step explanation:

Sophie can peel all potatoes in 45 minutes, in 1 minute she can peel:

1/45 of all the potatoes

Simon can peel all potatoes in 30 minutes, in 1 minute he can peel:

1/30 of all the potatoes

Together, in 1 minute they can peel:

1/45+1/30= 2/90+3/90= 5/90= 1/18 of all the potatoes

To peel all the potatoes they need:

1 ÷ 1/18= 1 × 18= 18 minutes

So it will take them 18 minutes to peel all the potatoes

7 0

3 years ago