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olga2289 [7]
4 years ago
12

If six people decide to come to a basketball game, but three of them are only 2/5 sure that they will stay for the entire time (

the other three are sure they'll stay the whole time), what is the probability that at the end, at least 5 people stayed the entire time
Mathematics
1 answer:
exis [7]4 years ago
7 0

Answer:

the probability that at the end, at least 5 people stayed the entire time = 0.352

Step-by-step explanation:

From the question, 3 of the people are sure to stay the whole time. So, we'll deduct 3 from 6.which leaves us with 3 that are only 2/5 or 0.4 sure that they will stay the whole time.

Thus, what we need to compute to fulfill the probability that at the end, at least 5 people stayed the entire time of which we know 3 will stay, so for the remaining 3,we'll compute;

P[≥2] which is x~bin(3,0.4)

Thus;

P(≥2) = (C(3,2) x 0.4² x 0.6) + (C(3,3) x 0.4³)

P(≥2) = 0.288 + 0.064

P(≥2) = 0.352

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Your question does not say what were your options, therefore I will answer generically: in order to understand if a point (ordered pair) is contained in a line, you need to substitute the x-component of the pair in the equation of the line and see if the calculations give you the y-component of the pair.

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y</span> = <span>4/3·0 + 1/3 = 1/3 </span>≠ 0
Therefore, the line does not contain (0, 0)

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Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut
frozen [14]

Answer:

a)\sqrt{144-288t+208t^2} b.) -12knots, 8 knots c) No e)4\sqrt{13}

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved 12\dot t (n.m) and ship B has moved 8\dot t (n.m). We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}

b)

We want to find \frac{ds}{dt} for t=0 and t=1

\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0

Since function f(x)=199-288x+208x^2 is quadratic, concave up and has no real roots, we know that 199-288x+208x^2>0 for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

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So, lets check this limit:

\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}

Notice that:

4\sqrt{13}=\sqrt{12^2+5^2}=√(speed of ship A² + speed of ship B²)

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Answer:

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