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:

$h_{ 1 } = 5 \cdot \ 12+11 = 71 \ inch \ \\ h_{ 2 } = 6 \cdot \ 12+3 = 75 \ inch \ \\ h_{ 3 } = 6 \cdot \ 12+6 = 78 \ inch \ \\ h_{ 4 } = 6 \cdot \ 12+2 = 74 \ inch \ \\ h_{ 5 } = 6 \cdot \ 12 = 72 \ inch \ \\ \ \\ b = \dfrac{ h_{ 1 }+h_{ 2 }+h_{ 3 }+h_{ 4 }+h_{ 5 } }{ 5 } = \dfrac{ 71+75+78+74+72 }{ 5 } = 74 \ inch \ \\ 74 = 12 \times 6 + 2 \ \\ \ \\ a = 6'2"$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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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  Tips to related online calculators
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