F(t)=256e(0.0611t) where t is in minutes. How long will it take the population to reach 5 times its initial value?

The distances in the Olympic final were farther on average.

Step-by-step explanation:

Let's consider the first answer choice:

The distances in the Olympic final were farther on average.

We can't calculate the mean values of each distribution, but we can see visually that the center of the Olympic final distribution is higher than the center of the U.S. qualifier distribution.

So we can say that the distances in the Olympic final were farther on average.

Now, let's consider the second answer choice:

The distances in the Olympic final were all greater than the U.S. qualifier distances.

Since the distributions overlap, there are some distances in the Olympic final distribution that are less than some of the distances in the U.S. qualifier distribution.

So we can't conclude that the distances in the Olympic final were all greater than the U.S. qualifier distances.

Lastly, let's consider the third answer choice:

The distances in the Olympic final varied noticeably more than the U.S qualifier distances.

We can see visually that the spread is close to same in each distribution, and if there is a difference, there appears to be more spread in the U.S. qualifier distribution. We can't calculate MAD from these displays, but we can see that the IQR of the U.S. qualifier distribution is larger.

So we can't say that the distances in the Olympic final varied noticeably more than the U.S qualifier distances