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3 years ago

Please help me with my homework!!! Number 4

Mathematics

1 answer:

grigory [225]3 years ago

7 0

Ill say about $
1125 over 5 years

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Instructions: Use the ratio of a 30-60-90 triangle to solve for the variables. Make sure to simplify radicals. Leave your answer

Jobisdone [24]

N = 50√3
m = 5√2
just use trigonometry

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3 years ago

Pls help with number one

LenaWriter [7]

A) You can drop hammer and a feather, both will reach ground at the same time ( it was actually done in Apollo mission)

b) i. Yes, it is similar to weightlessness experienced by astronauts

ii. The phenomenon of "weightlessness" occurs when there is no force of support on your body. When your body is effectively in "free fall" in roller coaster, accelerating downward at the acceleration of gravity, then you are not being supported.
Just like astronauts in space feel no force of support that we on earth feel due to gravity.

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3 years ago

35percent as a fraction in simplest form

Luden [163]

Answer:

7/20

Step-by-step explanation:

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3 years ago

Which number line correctly shows 4.5 - 2.5?

il63 [147K]

Answer:

number one

Step-by-step explanation:

if if 4.5 - 2.5 the answer is 2 that means the answer is number one

5 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

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

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