Question

evaluate (6 \times 0.222 \times 7.0) and round the answer appropriately. express your answer numerically using the appropriate number of significant figures. view available hint(s) submit previous answers incorrect; try again; 9 attempts remaining your answer implies that all the given values have at least four significant figures

Explanation:

Step1: Multiply the first two numbers

First, calculate \(6\times0.222\).
\(6\times0.222 = 1.332\)

Step2: Multiply the result by 7.0

Now, multiply \(1.332\) by \(7.0\).
\(1.332\times7.0=9.324\)

Step3: Determine significant figures

The number \(6\) has 1 significant figure, \(0.222\) has 3 significant figures, and \(7.0\) has 2 significant figures. When multiplying, the result should have the least number of significant figures among the factors, which is 1 from the number \(6\)? Wait, no, wait. Wait, \(6\) here - if \(6\) is considered as having 1 significant figure, \(0.222\) has 3, \(7.0\) has 2. But maybe \(6\) is an exact number (like 6 items, so infinite significant figures), \(0.222\) has 3, \(7.0\) has 2. So the limiting factor is 2 (from \(7.0\)) or 3 (from \(0.222\))? Wait, no, let's re - check. If \(6\) is a measured value with 1 significant figure, \(0.222\) (3), \(7.0\) (2). The rule for multiplication/division is that the result has the same number of significant figures as the least precise measurement. But if \(6\) is an exact integer (like a count), then we consider \(0.222\) (3 sig figs) and \(7.0\) (2 sig figs). So the least is 2? Wait, no, \(7.0\) has two significant figures (the trailing zero after the decimal is significant), \(0.222\) has three, and if \(6\) is a counted number (exact), then the number of significant figures is determined by the least among the measured values, which is \(7.0\) (2) or \(0.222\) (3)? Wait, no, the correct approach: when multiplying, the number of significant figures in the result is equal to the number of significant figures in the least precise measurement. Let's check the values:

- \(6\): If it's a measured value, 1 significant figure. If it's an exact value (like 6 molecules), then it has infinite significant figures.
- \(0.222\): 3 significant figures.
- \(7.0\): 2 significant figures (the zero after the decimal is significant because it's after a non - zero digit and the decimal is present).

Assuming \(6\) is an exact value (for example, if we are multiplying 6 times a measurement), then we look at \(0.222\) (3) and \(7.0\) (2). The least number of significant figures is 2? Wait, no, \(7.0\) has two, \(0.222\) has three. Wait, no, maybe I made a mistake. Let's calculate the product first: \(6\times0.222\times7.0 = 6\times1.554=9.324\). Now, let's check the significant figures:

If \(6\) is a measured value with 1 sig fig, \(0.222\) (3), \(7.0\) (2). The result should have 1 sig fig? But that would be 9. But that seems too rough. Wait, maybe \(6\) is considered as having 1 significant figure, \(0.222\) (3), \(7.0\) (2). But maybe the problem considers \(6\) as having 1, \(0.222\) as 3, \(7.0\) as 2, and the least is 1? But that gives 9. But maybe the \(6\) is an exact integer (like 6 times, so it's exact, so we can ignore its significant figures). Then we have \(0.222\) (3) and \(7.0\) (2). So the result should have 2 significant figures. So \(9.324\) rounded to 2 significant figures is \(9.3\)? Wait, no, 9.324 rounded to 2 significant figures: the first two significant figures are 9 and 3, the next digit is 2, which is less than 5, so we keep it as 9.3? Wait, no, 9.324 with two significant figures: 9.3 (since the first two are 9 and 3, the third digit is 2, so we don't round up). Wait, but let's check again. Wait, maybe the \(6\) is a measured value with 1 significant figure, \(0.222\) (3), \(7.0\) (2). The rule is that when multiplying, the number of significant figures is determined by the factor with the least number of significant figures. So if \(6\) has 1, then the result should have 1. But \(9.324\) with 1 significant figure is 9. But maybe the problem considers \(6\) as having 1, \(0.222\) as 3, \(7.0\) as 2, and the least is 1? But that seems odd. Wait, maybe I misread the numbers. The original problem is \(6\times0.222\times7.0\). Let's recalculate: \(6\times0.222 = 1.332\), \(1.332\times7.0=9.324\). Now, let's check the significant figures of each factor:

- \(6\): If it's a whole number without a decimal, it could be considered as having 1 significant figure (if it's a measured value) or infinite (if it's exact, like 6 objects).
- \(0.222\): 3 significant figures (the leading zero is not significant, the 2,2,2 are significant).
- \(7.0\): 2 significant figures (the zero after the decimal is significant).

If we take the least number of significant figures from the measured values (assuming \(6\) is measured with 1, \(0.222\) with 3, \(7.0\) with 2), the least is 1. But that would give 9. But maybe the \(6\) is exact, so we use the least from \(0.222\) (3) and \(7.0\) (2), so 2. So \(9.324\) rounded to 2 significant figures is \(9.3\)? Wait, no, 9.324 to two significant figures: the first two are 9 and 3, the next digit is 2, so we round down, so 9.3. But wait, maybe the \(6\) is considered to have 1 significant figure, \(0.222\) 3, \(7.0\) 2. The correct way is that when multiplying, the result's significant figures are equal to the minimum number of significant figures in the factors. So if \(6\) has 1, \(0.222\) has 3, \(7.0\) has 2, the minimum is 1. So the answer should be 9 (with 1 significant figure). But that seems too simple. Wait, maybe the problem has a typo, or maybe I misinterpret the significant figures. Wait, let's check the hint: "Your answer implies that all the given values have at least four significant figures". So the correct number of significant figures should be less than four. Let's see: \(6\) has 1, \(0.222\) has 3, \(7.0\) has 2. The product is \(9.324\). The least number of significant figures among the factors is 1 (from 6), but maybe \(6\) is considered as having 1, \(0.222\) as 3, \(7.0\) as 2, and we take the least, which is 1. But that would be 9. Alternatively, maybe \(6\) is an exact integer (so infinite sig figs), \(0.222\) (3), \(7.0\) (2). So the result should have 2 sig figs. So \(9.324\) rounded to 2 sig figs is \(9.3\). But let's check: \(6\times0.222 = 1.332\) (3 sig figs), then \(1.332\times7.0\): \(7.0\) has 2 sig figs, so the result should have 2 sig figs. So \(1.332\times7.0 = 9.324\approx9.3\) (2 sig figs). Or maybe \(6\) is considered to have 1 sig fig, \(0.222\) 3, \(7.0\) 2, and the least is 1, so 9. But the hint says the answer implies all have at least four, so we need to use the correct number of sig figs. Let's re - evaluate the significant figures:

- \(6\): If it's a measured value, 1 sig fig.
- \(0.222\): 3 sig figs.
- \(7.0\): 2 sig figs.

The rule for multiplication/division: the result has the same number of sig figs as the factor with the least number of sig figs. So the least is 1 (from 6), so the result should have 1 sig fig. So \(9.324\) rounded to 1 sig fig is 9. But maybe the \(6\) is an exact number (like 6.0, but it's written as 6). Wait, maybe the problem considers \(6\) as having 1 sig fig, \(0.222\) as 3, \(7.0\) as 2, and we take the minimum, which is 1. So the answer is 9.

Answer:

\(9\)