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2 years ago

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Consider the following statement. For every integer m, 7m 4 is not divisible by 7. Construct a proof for the statement by select

ing sentences from the following scrambled list and putting them in the correct order.

1 answer:

Maksim231197 [3]2 years ago

6 0

The constructed proof is given below.

<h3>Calculations and Parameters</h3>

Prove: f(M)= 7M +4 is not divisible by 7 for any integer M.

M=0 --> f(M) = 4 which is not divisible by 7.

M=1 --> f(M) = 11 which is not divisible by 7.

Suppose f(M) is not divisible by 7 for some positive integer M.

(this is the GIVEN induction hypothesis)

7(m+1)+4 = 7m+7+4 = 7m +4 + 7 = (7m+4)+7

dividing by 7, the quotient is [(7m+4)+7]/7

= (7m+4)/7 + 1

This is NOT divisible by 7 per induction hypothesis...

Read more about discrete mathematics here:

brainly.com/question/27793609

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company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul

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Sampling distribution involves the proportions of a data element in a given sample.

<em>The proportion of Good TV set is 0.67</em>
<em>The number of ways of selecting 5 from 6 TV sets is 6</em>
<em>The number of ways of selecting 4 from 6 TV sets is 15</em>

<em />

Given

$n = 6$

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

$Good = 4$

So, the proportion (p) that are good are:

$p = \frac{Good}{n}$

$p = \frac{4}{6}$

$p = 0.67$

<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 5$

So, we have:

$^6C_5 = \frac{6!}{(6 - 5)!5!}$

$^6C_5 = \frac{6!}{1!5!}$

$^6C_5 = \frac{6 \times 5!}{1 \times 5!}$

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<u>(c) All possible sample space of 4</u>

First, we calculate the number of ways to select 4.

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 4$

So, we have:

$^6C_4 = \frac{6!}{(6 - 4)!4!}$

$^6C_4 = \frac{6!}{2!4!}$

$^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}$

$^6C_4 = \frac{30}{2}$

$^6C_4 = 15$

So, the table is as follows:

$\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]$

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$p = \frac{Good}{n}$

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In (a), we have:

$p = 0.67$ --- proportion of Good TV

The sampling error is calculated as follows:

$SE_n = |p - p_n|$

So, we have:

$\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]$

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