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3 years ago

Um can someone help me with a question? (theres an attachment with the question there)

Mathematics

1 answer:

Ugo [173]3 years ago

4 0

She needs 15 cans of varnish

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Explain Well please.

Art [367]

The answer is C，
as she uses SSS to prove its congruent, she uses sides instead of angles so option a and d is not correct. as it needs to prove triangle pqr is congruent to triangle sqr, its not necessary to prove PQ is equal to QR. so the answer is c

7 0

3 years ago

Solve the polygon for x

wariber [46]

Octagon has sum of angles 1080°

$\\ \rm\Rrightarrow 5(10x+15)+2(14x)+6x-3=1080$

$\\ \rm\Rrightarrow 50x+75+28x+6x-3=1080$

$\\ \rm\Rrightarrow 84x+72=1080$

$\\ \rm\Rrightarrow 84x=1008$

$\\ \rm\Rrightarrow x=1008/84$

$\\ \rm\Rrightarrow x= 12$

4 0

2 years ago

3/4x+7/8=4 1/4 please help me

Orlov [11]

Answer:

4 1/2

Step-by-step explanation:

4 0

1 year ago

Is theres 10 birds and i shoot 1 how many is there left?

bazaltina [42]

There would be 9 birds left. 10-1=9

3 0

3 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago