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Mice21 [21]
3 years ago
9

the area of a parallelogram is 48cm².if the two adjacent sides are 8cm and 6cm, find the length of its diagonal .​

Mathematics
1 answer:
Travka [436]3 years ago
3 0

Answer:

10cm

Step-by-step explanation:

Assuming the shape is a rectangle since it's already stated that it's a parallelogram, and the area is stated, we can use the Pythagorean theorem to find the length

a^{2} +b^2=c^2, where c will be the length

isolate c

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

substitute for a and b

c=\sqrt{8^2+6^2}

c=\sqrt{64+36}

c=\sqrt{100}

c=10

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30 x 5/6 = 25

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I hope this helped. Let me know if you have any other questions!
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Is 15 over 45 equivalent to 45 over 135
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X² + 7x + 12 factorise it​
Sidana [21]

Step-by-step explanation:

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4 0
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
How would i solve for x and y in 3x+2y=17 and 2x-y=2 by using substitution
maxonik [38]
I’ll do an example problem, and I challenge you to do this on your own!
4x+6y=23
7y-8x=5
Solving for y in 4x+6y=23, we can separate the y by subtracting both sides by 4x (addition property of equality), resulting in 6y=23-4x. To make the y separate from everything else, we divide by 6, resulting in (23-4x)/6=y. To solve for x, we can do something similar - subtract 6y from both sides to get 23-6y=4x. Next, divide both sides by 4 to get (23-6y)/4=x.

Since we know that (23-4x)/6=y, we can plug that into 7y-8x=5, resulting in
7*(23-4x)/6-8x=5
= (161-28x)/6-8x
Multiplying both sides by 6, we get 161-28x-48x=30
= 161-76x
Subtracting 161 from both sides, we get -131=-76x. Next, we can divide both sides by -76 to separate the x and get x=131/76. Plugging that into 4x+6y=23, we get 4(131/76)+6y=23. Subtracting 4(131/76) from both sides, we get
6y=23-524/76. Lastly, we can divide both sides by 6 to get y=(23-524/76)/6

Good luck, and feel free to ask any questions!
4 0
3 years ago
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