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KiRa [710]
3 years ago
10

Plz help I’ll give brainlest

Mathematics
2 answers:
ankoles [38]3 years ago
5 0

Answer:

13 inches

Step-by-step explanation:

multiply 2 and 6.5 *i got 2 because the 2 hours and the 4 inches

In-s [12.5K]3 years ago
3 0

Answer:

about 13 inches sorry if im wrong ive spent so long tryna figure this out

Step-by-step explanation:

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The length of the rectangle is 4cm less than its width . what are the dimensions of the rectangle if its area is 140
melomori [17]
Let the width of the rectangle be x
Then length =x-4
Area=x(x-4)=140
x²-4x=140
x²-4x-140=0
x²-14x+10x-140=0
x(x-14)+10(x-14)=0
(x-14)(x+10)=0
Therefore,length=14 cm and  width=10 cm

8 0
3 years ago
Please help me!!!! I will give brainliest!
lisabon 2012 [21]
36=90/100 x a; 36=9/10 x a; a= 36/(9/10); a=360/9; a= 40

The Ice Cream shop served 40 customers.
5 0
3 years ago
Read 2 more answers
Factor out the coefficient of the variable -3.6m + 10.8
MA_775_DIABLO [31]

Answer: -1

Step-by-step explanation: First, factor out 3.6. You should get -m+3. Since the coefficient is just the number being multiplied to the variable (which is m here), your coefficient should be -1 because m is being multiplied to -1 to make -m. Hope this helps!

3 0
3 years ago
P is located at (8,-7) and Q is located at (-6,-3) on the coordinate plane
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Answer:

Step-by-step explanation:

1,-5

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
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