The heights of five starters on redwood high’s basketball team are 5’11”, 6’3”, 6’6”, 6’2” and 6’. The average of height of these players is?

Result

a = ft (Correct answer is: 6'2")

#### Solution:

$h_{ 1 } = 5 \cdot \ 12+11 = 71 \ inch \ \\ h_{ 2 } = 6 \cdot \ 12+3 = 75 \ inch \ \\ h_{ 3 } = 6 \cdot \ 12+6 = 78 \ inch \ \\ h_{ 4 } = 6 \cdot \ 12+2 = 74 \ inch \ \\ h_{ 5 } = 6 \cdot \ 12 = 72 \ inch \ \\ \ \\ b = \dfrac{ h_{ 1 }+h_{ 2 }+h_{ 3 }+h_{ 4 }+h_{ 5 } }{ 5 } = \dfrac{ 71+75+78+74+72 }{ 5 } = 74 \ inch \ \\ 74 = 12 \times 6 + 2 \ \\ \ \\ a = 6'2"$

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Dr Math
For conversion inches to feet and inch we user reminder calculator: https://www.hackmath.net/en/calculator/quotient-and-remainder?n=74&d=12 74 = 12 × 6 + 2 74 inch = 6 ft and 2 inch

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