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ira [324]
3 years ago
15

The perimeter of a rectangle is twice the sum of its length and its width. The perimeter is 22 meters and its length is 2 meters

more than twice its length
Mathematics
2 answers:
Nitella [24]3 years ago
7 0
Hello,
Let l represent length
w represent width
P represent perimeter

P=2(l+w)
22=2(l+w)
Divide 2 from both side
22/2=2(l+w)/2
11=l+w

l=2w+2
Apply this into the first equation
2w+2+w=11
3w+2=11
Subtract 2 from both side
3w+2-2=11-2
3w=9
Divided 3 from both side
3w/3=9/3
w=3 meters
l=2w+2
l=2(3)+2
l=8 meters
If you want to find the area
A=l×w
A=8×3
A=24 square meters. As a result, the length of this rectangle is 8 meters, width is 3 meters, and area is 24 square meters. Hope it help!
Leto [7]3 years ago
5 0
Perimeter=22cm
width=x
length=2x+2

2(3x+2)=22
6x+4=22
6x=18
x=3

width=3
length=8

area=3x8
=24m2
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C

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3 years ago
Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p
Ymorist [56]

Answer:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

A\sim Uniform(1 ,7)

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

B\sim Exp(\lambda=\frac{1}{30 min})

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

E(A +B)

Since the two random variables are assumed independent, then we have this

E(A+B) = E(A)+E(B)

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by E(X) =\frac{a+b}{2} where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

E(A)=\frac{1+7}{2}=4 hours

The expected value for the exponential distirbution is given by :

E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx

If we use the substitution y=\lambda x we have this:

E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}

Where X represent the random variable and \lambda the parameter. If we apply this formula to our case we got:

E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

And in order to find the variance for the random variable A+B we can find the individual variances:

Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2

Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2

We have the following property:

Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)

Since we have independnet variable the Cov(A,B)=0, so then:

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

3 0
3 years ago
6. The ABC Book Club charges a $40 monthly fee, plus $2 per book read in that month. The Easy Book Club charges a $35 monthly fe
d1i1m1o1n [39]

The expression for the cost of the ABC Book Club is 2<em>x </em>+ 40.  The expression for the cost of the Easy Book Club is 3<em>x</em> + 35.  To find when the total charge for both book clubs is equal, the two expressions must equal each other. (<em>x </em>= the number of books read)

2<em>x </em>+ 40 = 3<em>x</em> + 35

Subtract 35 from both sides.

2<em>x </em>+ 5 = 3x

Subtract 2<em>x </em>from both sides.

5 = <em>x</em>

So, the total charge for each club is equal when 5 books are read.

5 0
3 years ago
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