The ages of the members of a gym have a mean of 4747 years and a standard deviation of years. what can you conclude from chebyshev's theorem about the percentage of gym members aged between 17.317.3 and 76.776.7?

Chebyshev came up with the limits on how much or how many of the data must lie close to the mean. In specific for any positive k, the proportion of the data that lies within k standard deviations of the mean is at least: 

1 - 1/k² 

In this problem  the mean is 47 yrs therefore:

(47 – 17.3) = 29.7 = (76.7 - 47) 

The value of k is calculated using the formula:

29.7 / 11 = 2.7 = k

So the % of gym members aged between 19.4 and 76.6 is: 

1 - 1 / (2.6)² = 0.863 = 86.3 %

Therefore 86.3% of the gym members are aged between 19.4 and 76.6

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation: