Question 5 (1 point) The amount of sunlight a diver can see is halved for every $11 \mathrm{~m}$ she dives. What percentage of light remains when the diver is $60 \mathrm{~m}$ below the surface? a) $4.88 \%$ b) $2.28 \%$ c) $0.0228 \%$ d) $228 \%$

Step 1 ：The problem states that the amount of sunlight a diver can see is halved for every 11m she dives. This means that the remaining light can be calculated by halving the light for each 11m increment.

Step 2 ：This is a geometric progression problem where the common ratio is 0.5 (since the light is halved each time) and the number of terms is the depth divided by 11m.

Step 3 ：The formula for the nth term of a geometric progression is \(ar^{(n-1)}\), where a is the first term (100% in this case, since the diver starts with 100% of the light), r is the common ratio, and n is the number of terms.

Step 4 ：Substituting the given values into the formula, we get \(a = 100\), \(r = 0.5\), and \(n = 60/11 = 5.454545454545454\).

Step 5 ：Calculating the remaining light, we get \(remaining\_light = 100 * (0.5)^{5.454545454545454} = 4.560875330254521\).

Step 6 ：The remaining light is approximately 4.56%, which is closest to option a) 4.88%.

Step 7 ：Therefore, the final answer is \(\boxed{a) 4.88 \%}\).