A tech company bought a number of computers for its employees: 66 costing \$1600$1600 each, 44 costing \$1200$1200 each, and yy costing \$900$900 each, where yy is a positive odd integer. If the median price for all the computers purchased by the company was \$1200,$1200, what is the greatest possible value of yy

Answers

Answer 1

Answer:

5

Step-by-step explanation:

A tech company bought a number of computers for its employees: 6 costing $1600 each, 4 costing $1200 each, and y costing $900 each, where y is a positive odd integer. what is the greatest possible value of y

Solution:

The median of a group of numbers is the middle number when the group of numbers have been arranged either in ascending of descending other. The median of an even group of numbers is the average of the two most middle most numbers while the median of an odd group of numbers is the middle number.

Since there are 6 computers costing $1600, 4 computers costing $1200 each and y computers costing $900 each. Since y is odd, hence y + 6 + 4 would be odd and the median would be the middle number.

Given that the median price = $1200, this means that the middle of the group of numbers is $1200

900, 900, ..., 1200, 1200, 1200, 1200, 1600, 1600, 1600, 1600, 1600, 1600

Since y is odd, for the median price to be $1200, y has to be 3 or 5.

If y = 3:

900, 900, 900, 1200, 1200, 1200, 1200, 1600, 1600, 1600, 1600, 1600, 1600

median = $1200

If y = 5:

900, 900, 900, 900, 900, 1200, 1200, 1200, 1200, 1600, 1600, 1600, 1600, 1600, 1600

Median = $1200

Therefore the greatest possible value of y is 5

Answer 2

By knowing that the median must be $1,200, we will see that the greatest possible value of y is y = 9.

We know that the company bought:

6 computers costing $1,6004 computers costing $1,200y computers costing $900

We know that y is a positive odd number.

We also know that the median is $1200, now we want to get the maximum value of y such that the median is $1200. (Remember that the median is the value of the middle element of the set).

Then we will have the distribution:

{ y elements,  $1200, the other 3 $1200 computers plus the 6 $1600 computers}

Notice that y (the number of elements at the left of the median) must be equal to the number at the right of the median.

At the right of the median, we have a total of 9 computers, then y must be equal to 9.

The greatest possible value of y is 9.

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Answers

Answer:

[tex]f(x) = \left \{ {{39.99; 0 \le x \le 450} \atop {0.45x - 162.51; \ x>450}} \right.[/tex]

Step-by-step explanation:

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Open bracket

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[tex]f(x) = 39.99 + 0.45x - 202.5[/tex]

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