1) (3.4617 x 10^{7}) ÷ (5.61 x 10¯^{4})

The calculator shows 6.1706 x 10^{10} which then rounds to 6.17 x 10^{10} - three significant figures. The value which dictates this is in boldface.

2) [(9.714 x 10^{5}) (2.1482 x 10¯^{9})] ÷ [(4.1212) (3.7792 x 10¯^{5})]

The calculator shows 1.3398 x 10^{1} which then rounds to 13.40 - four significant figures. In this problem pay attention to order of operations, since division is not commutative. There are two ways to do this problem using the calculator: 1) multiply the last two numbers, put the result in memory, multiply the first two, then divide that by what is in memory or 2) multiply the first two numbers then do two divisions.

3) (4.7620 x 10¯^{15}) ÷ [(3.8529 x 10^{12}) (2.813 x 10¯^{7}) (9.50)]

The calculator shows 4.625 x 10¯^{22}, which then rounds to 4.62 x 10¯^{22} - three significant figures. Notice the use of the rounding with five rule.

4) [(561.0) (34,908) (23.0)] ÷ [(21.888) (75.2) (120.00)]

The calculator shows 2280.3972, which rounds off to 2280, three significant figures. In scientific notation, this answer would be 2.28 x 10^{3}.

Note this last use of scientific notation to indicate significant figures where otherwise you might not realize they were significant. For example, 2300 looks like it has only two significant figures, but you know (from the problem) it really has three. How do you show this. One way is to use scientific notation like this: 2.30 x 10^{3}. Now the 2.30 portion clearly has three significant figures.
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