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3 years ago

I need help on this question please

Mathematics

1 answer:

Fynjy0 [20]3 years ago

5 0

Answer: NEITHER

Step-by-step explanation:

From the equation of lines given above, to know if relationships exist between the two lines, we must first determine that from their gradients or slopes.

From first equation

2x - y = 5,

y = 2x - 5

Therefore, m₁ = 2

From the second equation

3x - y = 5

y = 3x - 5

Therefore m₂ = 3

Recall, For the two lines to be parallel, m₁ = m₂ , and for the two line to be perpendicular, m₁m₂ = -1

since none fulfilled the conditions stated above, the answer then is

NEITHER

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Simplify -2+3(1-4)-2

oksian1 [2.3K]

Answer:

-13

Step-by-step explanation:

−2+3(1−4)−2

=−2+(3)(−3)−2

=−2+−9−2

=−11−2

=−13

hope it helped

please mark me as brainliest.

4 0

3 years ago

Yo help 20 points plssssssssssssss

Rina8888 [55]

Answer:

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6 0

2 years ago

A red light flashes every 14 minutes. A blue light flashes every 24 minutes. when will the two lights flash together again if th

natta225 [31]

14×12=168

24×7=168

168÷60=2 hours and 48 minutes

2 hours and 48 minutes+ 8:00AM=10:48AM

Answer: 10:48AM

4 0

2 years ago

Answer the question right or I report you

zubka84 [21]

Answer:

3 ÷ 0.25

Step-by-step explanation:

There are three squares, so 3.

Each square is divided into four squares,

One square out of four squares is,

1 / 4 = 0.25

Therefore,

3 ÷ 0.25

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

2 years ago