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Nostrana [21]
3 years ago
9

List all the elements of the sample space for the following experiment: You spin a spinner with four equal sections labeled 1, 2

, 3, and 4 and toss a dime. a. (1, H), (2, H), (3, H), (4, H), (1, H), (2, H), (3, H), (4, H) b. (1, H), (2, H), (3, H), (4, H), (1, T), (2, T), (3, T), (4, T) c. (1, T), (2, T), (3, T), (4, T), (1, T), (2, T), (3, T), (4, T) d. (1, H), (H, 1), (2, H), (H, 2), (3, H), (H, 3), (4, H), (H, 4), (1, T), (2, T), (3, T), (4, T)
Mathematics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

Option B.

Step-by-step explanation:

When we spin a spinner the sample space for a spinner having four equal sections labeled 1, 2, 3, 4 will be {1, 2, 3, 4} and for tossing the dime is {H, T}.

Now if we toss a dime and spinning the spin is done simultaneously then sample space will have the elements {(1, H), (2, H), (3, H), (4, H), (1, T), (2, T), (3,T), (4, T)}

Therefore Option B. is the correct option.

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Among all right circular cones with a slant height of 24​, what are the dimensions​ (radius and​ height) that maximize the volum
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Answer:

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Step-by-step explanation:

We need to use the Pythagorean theorem to solve the problem.

The theorem indicates that,

r^2+h^2=24^2 \\r^2+h^2=576\\r^2=576-h^2

Once this is defined, we proceed to define the volume of a cone,

v=\frac{1}{3}\pi r^2 h

Substituting,

v=\frac{1}{3} \pi (576-h^2)h\\v=\frac{1}{3} \pi (576h-h^3)

We need to find the maximum height, so we proceed to calculate h, by means of its derivative and equalizing 0,

\frac{dv}{dh} = \frac{1}{3} \pi (576-3h^2)

\frac{dv}{dh} = 0 then \rightarrow \frac{1}{3}\pi(576-3h^2)=0

h_1=-8\sqrt{3}\\h_2=8\sqrt{3}

<em>We select the positiv value.</em>

We have then,

r^2 = 576-(8\sqrt3)^2 = 384\\r=\sqrt{384}

We can now calculate the maximum volume,

V_{max}= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\sqrt{384})^2 (8\sqrt{3}) = 5571.99

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Step-by-step explanation:

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