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djyliett [7]
3 years ago
7

(k/3^k)(x-6)^k

Mathematics
1 answer:
enot [183]3 years ago
5 0

\displaystyle\sum_{k\ge0}\frac k{3^k}(x-6)^k

converges by the ratio test for

\displaystyle\lim_{k\to\infty}\left|\frac{\frac{k+1}{3^{k+1}}(x-6)^{k+1}}{\frac k{3^k}(x-6)^k}\right|=\frac{|x-6|}3\lim_{k\to\infty}\frac{k+1}k

The limit is 1, so the series converges as long as

\dfrac{|x-6|}3

which indicates a radius of convergence of 3.

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Which of the following fractions is nearest to 1/2
Zinaida [17]

Answer:

12/25

Step-by-step explanation:

1/2 = 25/50

1) 3/5 = 30/50

2) 12/25 = 24/50  (closest to 25/50)

3) 27/50

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Select the expression that represents the following statement multiply 6 by 2, and then subtract 4. 2 - 4x6
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The answer is the second one, 6x2 - 4
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Help with #12, 13, 14, 15! I will mark u as brainliest answer
son4ous [18]
12) LCM of 2 & 12 is 12 
Answer: (F)

13) (25 x 8) x 4
Step 1: Order of operations do parenthesis first: (25 x 8) = 200
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7 0
3 years ago
Which of the following equations will yield the volume of a triangular prism?
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Answer:

V = Bh

Step-by-step explanation:

The equations will yield the volume of a triangular prism is V = Bh.

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B = Base

h = Height

<u>Here are an example : </u>

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2 years ago
A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% hi
Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

                = [0.20]⁴

                = 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (L)

               = 4 × (0.20)³ × 0.50

               = 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (M)

               = 4 × (0.20)³ × 0.30

               = 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = {4\choose 2} × [P (H)]² × [P (M)]²

                 = 6 × (0.20)² × (0.30)²

                 = 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

                                                            + P (HHMM)

                                                  = 0.0016 + 0.016 + 0.0096 + 0.0216

                                                  = 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

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3 years ago
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