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Sindrei [870]
3 years ago
11

The random numbers below represent 15 trials of a basketball simulation conducted using a spinner numbered 0–8. 76645, 46757, 28

334, 81357, 52453, 21761, 51537, 62385, 62135, 16687, 41662, 27135, 45445, 33858, 86427 Let the number 1 represent a 3-point basket and the numbers 2–8 represent 2-point baskets. Based on the simulation, what is the probability that at least one of the next 5 baskets made by the team is a 3-pointer? A. 6/15 B. 8/15 C. 5/15 D. 7/15
Mathematics
1 answer:
mote1985 [20]3 years ago
4 0

Answer: Option D.

Step-by-step explanation:

Out of the 15 simulations, the number with a 1 in them is 7. (are the sets of 5 shots that have at least one 3-pointer)

So out of 15 sets, 7 of them have at least one 3-point in them.

the experimental probability is p = 7/15.

So the correct option is D.

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4/5 of a number is more than 3/4 of the number by 5.find the numbers
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Use the surface integral in​ Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus
Alinara [238K]

Answer:

The circulation of the field f(x) over curve C is Zero

Step-by-step explanation:

The function f(x)=(x^{2},4x,z^{2}) and curve C is ellipse of equation

16x^{2} + 4y^{2} = 3

Theory: Stokes Theorem is given by:

I= \int \int\limits {{Curl f\cdot \hat{N }} \, dx

Where, Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]

Also, f(x) = (F1,F2,F3)

\hat{N} = grad(g(x))

Using Stokes Theorem,

Surface is given by g(x) = 16x^{2} + 4y^{2} - 3

Therefore, tex]\hat{N} = grad(g(x))[/tex]

\hat{N} = grad(16x^{2} + 4y^{2} - 3)

\hat{N} = (32x,8y,0)

Now,  f(x)=(x^{2},4x,z^{2})

Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]

Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\x^{2}&4x&z^{2}\end{array}\right]

Curl f(x) = (0,0,4)

Putting all values in Stokes Theorem,

I= \int \int\limits {Curl f\cdot \hat{N} } \, dx

I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx

I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx

I=0

Thus, The circulation of the field f(x) over curve C is Zero

3 0
3 years ago
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