An auctioneer sold a herd of cattle whose minimum weight was 920 pounds, median was 1160 pounds, standard deviation 76, and IQR 96 pounds. They sold for 50 cents a pound, and the auctioneer took a $15 commission on each animal. Then, for example, a steer weighing 1100 pounds would net the owner 0.50(1100)−15=$535. Find the minimum, median, standard deviation, and IQR of the net sale prices.

minimum = $445

median = $565

standard deviation = $38

IQR = $48

Step-by-step explanation:

The minimum and the median are measures of location and are affected during addition (or subtraction) and multiplication (or division).

Satandard deviation and IQR (inter-quartile range), which are measures of dispersion, are affected by only multiplication (or division).

For the weight,

minimum = 920 pounds

median = 1160 pounds

standard deviation = 76 pounds

IQR = 96 pounds

For the price,

minimum = 0.50(920) - 15 = $445

median = 0.50(1160) - 15 = $565

standard deviation = 0.50(76) = $38(Subtraction discarded)

IQR = 0.50(96) = $48(Subtraction discarded)

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0