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lara [203]
3 years ago
5

A plane flying horizontally at an altitude of 3 mi and a speed of 440 mi/h passes directly over a radar station. find the rate a

t which the distance from the plane to the station is increasing when it is 4 mi away from the station. (round your answer to the nearest whole number.)
Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
8 0
Let α represent the acute angle between the horizontal and the straight line from the plane to the station. If the 4-mile measure is the straight-line distance from the plane to the station, then
  sin(α) = 3/4
and
  cos(α) = √(1 - (3/4)²) = (√7)/4

The distance from the station to the plane is increasing at a rate that is the plane's speed multiplied by the cosine of the angle α. Hence the plane–station distance is increasing at the rate of
  (440 mph)×(√7)/4 ≈ 291 mph
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faust18 [17]

I'll use t for the messages Teresa sent, e for messages Eric sent, and r for messages Ryan sent.

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From the text we also know:

e=3r ; r=t+6

If we substitute in the original equation we get:

t+3(t+6)+t+6=99; By simplifying it we get

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8 0
3 years ago
Suppose that the probability that any particle emitted by a radioactive material will penetrate a certain shield is 0.01. If 10
AnnyKZ [126]

Answer:

a)P=0.42

b) n\geq 297

Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

P=\begin{pmatrix}n\\ k\end{pmatrix} p^k(1-p)^{n-k}=\frac{n!}{k!(n!-k!)}p^k(1-p)^{n-k}

a) we have k=2, n=10 and p=0.01:

P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42

b) We have, 1-(1-p)^n=P, Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.

1-0.99^n\geq 0.95\\0.99^n\leq 1-0.95\\0.99^n\leq 0.05\\n\geq 297

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3 years ago
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aalyn [17]

Answer:

We are observing the galactic center as it was 27,000 years ago

Step-by-step explanation:

The Galactic Center, or Galactic Centre, is the rotational center of the Milky Way. It is 8,122 ± 31 parsecs (26,490 ± 100 ly) away from Earth in the direction of the constellations Sagittarius, Ophiuchus, and Scorpius where the Milky Way appears brightest. It coincides with the compact radio source Sagittarius A*

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