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Ilia_Sergeevich [38]
3 years ago
12

For the spinner above, the probability of landing on black is 1/2 and the probability of landing on red is 1/3 . Suppose it is s

pun three times. What is the probability it will land on white all three times?

Mathematics
2 answers:
blsea [12.9K]3 years ago
5 0

Answer:

The probability of it landing on white three consecutive times is 1/216

Step-by-step explanation:

In order to solve this problem we first need to calculate the probability of the spinner landing on white, we know that the sum of all three probablities must be equal to 1, so we subtract the given probabilities from 1 and that will be the probability of it landing on white three times. We have:

white = 1 - 1/3 - 1/2 = 1 - (2 + 3)/6 = 6/6 - 5/6 = 1/6

The probability of the spinner landing on white all three times is given by the probability of it landing on white one time powered by the number of times that it has to land on that collor. So we have:

white three = white^(3) = (1/6)^3 = 1/216

marin [14]3 years ago
4 0

Answer:

1/216

Step-by-step explanation:

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Marianna [84]

Answer:

13

Step-by-step explanation:

x² = 5²+12²

x² = 25+144

x² = 169

x² = 13²

x = 13

3 0
3 years ago
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r
Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt

If we ask that f has continuous (n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists c between a and x such that

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x

Hence,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .

Thus,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

and the Taylor theorem with Lagrange remainder is

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Step-by-step explanation:

5 0
3 years ago
A communications tower is supported by two wires, connected at the same point on the ground. One is attached to the tower at D a
viktelen [127]

Answer:

10.99 m

Step-by-step explanation:

✍️What we are basically asked to solve here is to find the distance between C and D.

To find CD, find the length of BC, and BD. Their difference will give us CD.

Thus, BC - BD = CD.

✍️Finding BC using trigonometric ratio formula:

\theta = 30 + 10 = 40

Opposite side = BC = ??

Adjacent side = 42 m

Thus:

tan(\theta) = \frac{opposite}{adjacent}

Plug in the values

tan(40) = \frac{BC}{42}

Multiply both sides by 42

tan(40) \times 42 = \frac{BC}{42} \times 42

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✍️Finding BD using trigonometric ratio formula:

\theta = 30

Opposite side = BD = ??

Adjacent side = 42 m

Thus:

tan(\theta) = \frac{opposite}{adjacent}

Plug in the values

tan(30) = \frac{BD}{42}

Multiply both sides by 42

tan(30) \times 42 = \frac{BD}{42} \times 42

24.25 = BD

BD = 24.25 m

✍️How much below the top of the tower is the shorter one attached:

Thus,

BC - BD = CD

35.24 m - 24.25 m = 10.99 m

7 0
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Which of the following are exterior angles? check all that apply
Zigmanuir [339]

Answer:

4, 5, 6

Step-by-step explanation:

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3 0
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podryga [215]

Answer:

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4 0
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