Find the value of x

Mathematics

1 answer:

iVinArrow [24]3 years ago

4 0

Answer:

x = 121

Step-by-step explanation:

Let the measure of the altitude of the given triangle be h units.

By geometric mean theorem:

$h = \sqrt{23 \times x} \\ \\ h = \sqrt{23x} \\ \\ now \: by \: pythagors \: theorem \\ \\ {(x + 11)}^{2} = {h}^{2} + {x}^{2} \\ \\ \cancel{{x}^{2}} + 22x + 121 = {( \sqrt{23x} )}^{2} + \cancel{{x}^{2}} \\ \\ 22x + 121 = 23x \\ \\ 22x - 23x = - 121 \\ \\ - x = - 121 \\ \\ x = 121$

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What is 3:120 equivalent to

fenix001 [56]

1:40 because if we think of this just like a fraction $\frac{3}{120}$ that simplifys down to $\frac{1}{40}$ and because fraction simplifications are similar (related to each other) we can concur that these would be too.

Enjoy!=)

6 0

3 years ago

Read 2 more answers

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

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