How many ants does it take to have the same mass as an elephant

First we observe that the mass of the ant is in grams, where the mass of the elephant is in metric tons. We cannot compare the two masses as is, so we first convert them into the same units. We can do this with ratios, using the conversion chart above to convert metric tons into kilograms and then kilograms into grams.

(8 metric tons)(1000 kg1 metric ton)(1000 grams1 kg)=8,000,000 grams=8×106 grams
Notice that both metric tons and kg appear once on both the top and the bottom, enabling them to be canceled out, leaving only grams. Also note that we could have converted .003 grams into metric tons, and though this would have given an answer with a large number of decimal places, we would end up with the same final result.

Now that both quantities are in grams, we want to find how many ants, n, are required to have the same mass as the elephant. As each ant has a mass of 4×10−3 grams, we multiply 4×10−3 by n to get the total mass of the ants
total mass of n ants=4×10−3n
and then we set this equal to the mass of the elephant:
4×10−3n=8×106
We find that n=2×109 ants.

Thus, it takes approximately 2×109 ants to have the same mass as an elephant.
Each ant is 10−1 cm long, and we have 2×109 ants total. If we put all these ants in a line (front to back), we can find the length of the line by multiplying the length of one ant by how many ants we have:
2×109×10−1 cm=2×108 cm
As is, this answer is difficult to illustrate because it is such a large number. To find two objects that are this distance apart, we first want to convert centimeters into more useful units. We do this by using ratios from the above chart, converting centimeters to meters and meters to kilometers.
(2×108 cm)(1 m102 cm)(1 km103 m)=2×103 km=2000 km
This amount is much easier to visualize, and it turns out that 2000 km is approximately the driving distance from San Francisco, California to Denver, Colorado.

The probability of picking 2 green markers is 2/10