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Goshia [24]
3 years ago
11

How to solve this question #4 ?

Mathematics
2 answers:
vagabundo [1.1K]3 years ago
7 0
I think it would be D.  4500 pi cm cubed
gayaneshka [121]3 years ago
5 0
This question is a piece-o-cake if you know the formulas for the area and volume of a sphere, and impossible of you don't.

Area of a sphere = 4 π R²  (just happens to be the area of 4 great circles)

Volume of a sphere = (4/3) π R³

We know the area of this sphere's great circle, so we can use the
first formula to find the sphere's radius.  Then, once we know the
radius, we can use the second formula to find its volume.

Area of 4 great circles = 4 π R²

Area of ONE great circle = π R²

           225 π cm²  =  π R²

               R²  =  225 cm²

               R  =  √225cm²  =  15 cm .

Now we have a number for R, so off we go to the formula for volume.

         Volume  =  (4/3) π R³

                      =  (4/3) π (15 cm)³

                      =  (4/3) π (3,375 cm³)

                       =      14,137.17  cm³      (rounded)

This answer feels very good UNTIL you look at the choices.
_____________________________________________________

I've gone around several loops and twists trying to find out what gives here,
but have come up dry.

The only thing I found is the possibility of a misprint in the question:  

If the area of a great circle is 225π cm², then the sphere's AREA is 900π cm².

I'm sure this is not the discrepancy.  I'll leave my solution here, and hope
someone else can find why I'm so mismatched with the choices.
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Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

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Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

No-show rate of 10%.

This means that p = 0.1

Four travelers who have made hotel reservations in this study.

This means that n = 4

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486

P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523

0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) What is the most likely value for X?

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561

P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916

P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486

P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

X = 0 has the highest probability, which means that 0 is the most likely value for X.

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Answer:

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