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3 years ago

9

An engineer earns $6000 per month. A contractor makes $80 per hour, while a teacher gets basic salary of $3000 and $50 per hour

any additional job. Considering a month, which job
is better;when, and why? Justify your answer using graphs,equations, and tables

1 answer:

kumpel [21]3 years ago

3 0

The contractor has a better job because he receives the highest salary among the three.

Multiplying 8 hours of work by 20 working days in a month, we will get 160 monthly working hours.

Now let's compute:
Engineer: $6000 per mo. / 160 hrs. = $37.5 per hr.
Contractor: $80 per hr. * 160 hrs. = $12,800 per mo.
Teacher: $3000 per mo. / 160 hrs. = $18.75 per hr. + $50 per hr = $68.75 per hr.

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(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

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Step-by-step explanation:

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Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

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Next, we list out all number of heads in each roll (sorted)

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Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

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Question 10: Who is likely to get number 5

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$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

<em>From the above calculations: </em> $0.267 > 0.03125$ <em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

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This means that the median is the mean of the 16th and the 17th item

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$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

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This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

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$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

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$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

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$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

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