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MaRussiya [10]
3 years ago
14

Let P (n) be the statement that a postage of n cents can be formed using just 4-cent stamps and 7-cent stamps. The goal is to sh

ow, using strong induction, that P (n) is true for all integers n ≥ 18.
Mathematics
1 answer:
natulia [17]3 years ago
7 0

Complete Question

Let P (n) be the statement that a postage of n cents can be formed using just 4-cent stamps and 7-cent stamps. The parts of this exercise outline a strong induction proof that P (n) is true for n ≥ 18.

Show statements P (18), P (19), P (20), and P (21) are true, completing the basis step of the proof.

Answer:

P(18) is true

P(19) is true

P(20) is true

P(21) is true

Step-by-step explanation:

a. When n = 18

18 cents can be formed using two 7cents and one 4cents

i.e. 2 * 7 + 4 = 18

So, P(18) is true

b. When n = 19

19 cents can be formed using one 7cents and three 4cents

i.e. 1 * 7 + 3 * 4 = 19

So, P(19) is true

c. When n = 20

18 cents can be formed using five 4cents

i.e. 5 * 4 = 20

So, P(20) is true

d. When n = 21

18 cents can be formed using three 7cents

i.e. 3 * 7 = 21

So, P(21) is true

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A rectangles length is 2 in more than its width. The area of the rectangle is 143 square inches. Find the dimensions of the rect
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Answer: The length is 13 inches and the width is 11 inches.

Step-by-step explanation:

Let L represent the length of the rectangle.

Let W represent the width of the rectangle.

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Substituting L = W + 2 into equation 1, it becomes

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4 0
3 years ago
Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam
tigry1 [53]

Answer:

P(A) = \frac{30}{100}

P(B) = \frac{77}{100}

P(A\ n\ B) = \frac{22}{100}

P(A\ u\ B) = \frac{85}{100}

Step-by-step explanation:

Given

See attachment for proper format of table

n = 100 --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

Yes = 22 and No = 8

So, we have:

n(A) = Yes + No

n(A) = 22 + 8

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P(A) is then calculated as:

P(A) = \frac{n(A)}{Sample}

P(A) = \frac{30}{100}

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

(1) = 22    (2) = 25 and (3) = 30

So, we have:

n(B) = (1) + (2) + (3)

n(B) = 22 + 25 + 30

n(B) = 77

P(B) is then calculated as:

P(B) = \frac{n(B)}{Sample}

P(B) = \frac{77}{100}

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

n(A\ n\ B) = 22

The probability is then calculated as:

P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}

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This gives:

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Take LCM

P(A\ u\ B) = \frac{30+77-22}{100}

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<h3>How to calculate the numbers?</h3>

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