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meriva
3 years ago
15

A sphere has a diameter of 12 ft. What is the volume of the sphere? Give the exact value in terms of .

Mathematics
2 answers:
Alchen [17]3 years ago
5 0

Answer:

904.78 is the answer

volume = 4/3 π r cubed

UNO [17]3 years ago
3 0

Answer:

288π

That is exact value in terms of pi.

Step-by-step explanation:

So if the diameter is 12, the radius is 6.

The volume of a sphere is \frac{4}{3} \pi r^3

Since r = 6, the volume is \frac{4}{3} * 216\pi

Which is 288π

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bixtya [17]

Answer:

the answer is 2880

Step-by-step explanation:

because you would use the formula (n - 2) x 180 which equals the sum interior angles of any polygon .

The letter n stands for number of sides so in this case you would do (18 - 2) x 180 so 16 × 180 = 2880

hope this helped :) pls ask if you need any more explanation .

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1 year ago
One-quarter of a pound of bananas costs $0.82. How much does one pound of bananas cost?
aev [14]

Answer:

.82 x 4 = $3.28

.82 = 1/4 pound, so you multiply by 4 (the reciprocal of 1/4) to get a whole.

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4 years ago
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Ksivusya [100]
Unit Form Is

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Step-by-step explanation:

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3 years ago
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(a) A parachutist lands at a point on the line between the points A and B, and the target is an operation at A. The operation fa
mr Goodwill [35]

Answer:

a) \frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}

b) P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736

Step-by-step explanation:

Part a

We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform Y\sim Unif(A,B). And the density function would be given by:

f(x) =\frac{1}{B-A} , A

And 0 for other case.

The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.

So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:

P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}

And we can find the probability desired like this:

P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)

And from the cumulative distribution function of X ficen by F(X)\frac{X-A}{B-A} we got:

\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}

Part b

For this case we assume that X\sim Gamma (2,1)

On this case we assume that \alpha=2, \beta= 1

The density function for the Gamma distribution is given by:

P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}

And on this case we can find the probability using the complement rule like this:

P(X>1) = 1-P(X\leq 1)=0.736

We can solve this problem with the following excel code:

"=1-GAMMA.DIST(1;2;1;TRUE)"

And if we do it by hand we need to do this:

P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736

6 0
3 years ago
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