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Semmy [17]
3 years ago
14

Bill is able to save $35/week after working part-time and paying his expenses. These two formulas show his weekly savings: f(1)

= 35, f(n) = f(1) + f(n-1) for n > 1 f(n) = 35n Which one of these formulas show the sequence written rercursively and which shows it written explicity? Justify your explanantions. Use the recursive formula to make a table of values for 1 ≤ n ≤ 5. Show your calculations. Use the explicit formula to demonstrate the most direct method to find f(40). Explain why you chose that method and what your answer means. Show your calculations. Given the sequence of numbers: 5, 6, 8, 11, 15, 20, 26, 33, 41,… Explain whether or not this sequence can be considered a function.
Mathematics
1 answer:
Umnica [9.8K]3 years ago
4 0
1. We use the recursive formula to make the table of values:
f(1) = 35
f(2) = f(1) + f(2-1) = f(1) + f(1) = 35 + 35 = 70
f(3) = f(1) + f(3-1) = f(1) + f(2) = 35 + 70 = 105
f(4) = f(1) + f(4-1) = f(1) + f(3) = 35 + 105 = 140
f(5) = f(1) + f(5-1) = f(1) + f(4) = 35 + 140 = 175

2. We observe that the pattern is that for each increase of n by 1, the value of f(n) increases by 35. The explicit equation would be that f(n) = 35n. This fits with the description that Bill saves up $35 each week, thus meaning that he adds $35 to the previous week's value.

3. Therefore, the value of f(40) = 35*40 = 1400. This is easier than having to calculate each value from f(1) up to f(39) individually. The answer of 1400 means that Bill will have saved up $1400 after 40 weeks.

4. For the sequence of 5, 6, 8, 11, 15, 20, 26, 33, 41...
The first-order differences between each pair of terms is: 1, 2, 3, 4, 5, 6, 7, 8...since these differences form a linear equation, this sequence can be expressed as a quadratic equation. Since quadratics are functions (they do not have repeating values of the x-coordinate), therefore, this sequence can also be considered a function.
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Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

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A. 5x - 1

Step-by-step explanation:

  3x - 2

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5x-1

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