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Veseljchak [2.6K]
3 years ago
10

A small accounting firm has 444 accountants who each earn a different salary between \$50{,}000$50,000dollar sign, 50, comma, 00

0 and \$60{,}000$60,000dollar sign, 60, comma, 000, and a 5^{\text{th}}5 th 5, start superscript, start text, t, h, end text, end superscript accountant who works part-time for tax season and earns \$10{,}000$10,000dollar sign, 10, comma, 000. [Show data] 1010 5252 5454 5656 5858 The firm decides to get rid of the part-time accountant and keep the other 444 salaries the same. How will getting rid of the part-time accountant affect the mean and median? Choose 1 answer: Choose 1 answer: (Choice A) A Both the mean and median will increase, but the median will increase by more than the mean. (Choice B) B Both the mean and median will increase, but the mean will increase by more than the median. (Choice C) C The mean will increase, and the median will decrease. (Choice D) D The median will increase, and the mean will decrease.
Mathematics
1 answer:
natta225 [31]3 years ago
4 0

Answer:

The correct option is;

(Choice B) Both the mean and median will increase, but the mean will increase by more than the median.

Step-by-step explanation:

Here, we are expected to show the differences (advantages/properties) of the mean and median

Note that the definition of the mean is the sum of the value of all data points divided by the number of data points

Assuming the four accountants each earn $ 55,000.00

Therefore, initial mean before getting rd of the part-time accountant is

(55+ 55+ 55 + 55 + 10)/5 = 46

Final mean after the part-time accountant is gone =(55+ 55+ 55 + 55 )/4 = 55

Therefore the mean increases

However, the median is given by the middle value when the list of values are arranged in increasing order as

55, 55, 55, 55, 10

Therefore the initial median is 55

Final median after the part-time accountant is gone is

55, 55, 55, 55

Therefore, final = (55 + 55)/2 = 55

That is the median remains the same that is little or increase.

Therefore, since the range of values for the full time accountants salaries weigh less than the difference between the average full time salary and a single part time, both the mean and the median will increase but the mean will increase more than the median

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kow [346]

The value of h(t) when t=\frac{15}{32} is 10.02.

Solution:

Given function h(t)=-16t^2+15t+6.5

To find the value of h(t) when t=\frac{15}{32}:

h(t)=-16t^2+15t+6.5

Substitute t=\frac{15}{32} in the given function.

$h\left(\frac{15}{32} \right)=-16\left(\frac{15}{32} \right)^2+15\left(\frac{15}{32} \right)+6.5

            $=-16\left(\frac{225}{1024} \right)+15\left(\frac{15}{32} \right)+6.5

Now multiply the common terms into inside the bracket.

           $=-\left(\frac{3600}{1024} \right)+\left(\frac{225}{32} \right)+6.5

Now, in the first term, the numerator and denominator both have common factor 16. So reduce the first term into the lowest term.

          $=-\left(\frac{225}{64} \right)+\left(\frac{225}{32} \right)+6.5

To make the denominator same, take LCM of the denominators.

LCM of 64 and 32 = 64

        $=-\left(\frac{225}{64} \right)+\left(\frac{225\times2}{32\times2} \right)+6.5\times\frac{64}{64}

        $=-\frac{225}{64} +\frac{450}{64}+\frac{416}{64}

        $=\frac{-225+450+416}{64}

       $=\frac{641}{64}

       = 10.02

$h\left(\frac{15}{32} \right)=10.02

Hence the value of h(t) when t=\frac{15}{32} is 10.02.

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