Question

2nd attempt
see periodic table
see hint
perform the following calculation and express the result to the correct number of significant figures. you can assume all the numbers are measurements.
(9.42 + 3.69) × 95.211 =
1st attempt

Explanation:

Step1: Perform the addition inside the parentheses

First, we calculate \(9.42 + 3.69\). When adding, we consider the decimal places. \(9.42\) has two decimal places and \(3.69\) has two decimal places.
\(9.42 + 3.69 = 13.11\)
But we need to consider significant figures for addition. The number of decimal places in the result should match the least number of decimal places in the numbers being added. Both \(9.42\) and \(3.69\) have two decimal places, so the sum \(13.11\) is appropriate for now (we will consider significant figures later when multiplying).

Step2: Multiply the result by 95.211

Now we multiply \(13.11\) by \(95.211\).
\(13.11\times95.211 = 13.11\times95.211\)
First, calculate the product: \(13.11\times95.211 = 1248.21621\)

Now, we need to consider significant figures for multiplication/division. The number of significant figures in the result should match the least number of significant figures in the numbers being multiplied.

- \(9.42\) has 3 significant figures.
- \(3.69\) has 3 significant figures. So their sum (when considering significant figures for addition, the sum \(13.11\) actually has 4 significant figures, but the limiting factor here is the multiplication step. Wait, no: when adding \(9.42\) (3 sig figs, two decimal places) and \(3.69\) (3 sig figs, two decimal places), the sum is \(13.11\) (which is 4 digits, but the decimal places are two, so the precision is to the hundredth. However, for significant figures in multiplication, we look at the number of significant figures in each factor.

Wait, let's re - evaluate:

The rule for addition/subtraction: the result has the same number of decimal places as the number with the least number of decimal places. \(9.42\) has 2 decimal places, \(3.69\) has 2 decimal places, so \(9.42 + 3.69=13.11\) (2 decimal places, 4 significant figures).

For multiplication/division: the result has the same number of significant figures as the number with the least number of significant figures.

- The first factor after addition: \(13.11\) has 4 significant figures.
- The second factor: \(95.211\) has 5 significant figures.

But wait, the original numbers in the addition: \(9.42\) (3 sig figs) and \(3.69\) (3 sig figs). When we add them, the sum's precision is to the hundredth, but the number of significant figures: \(13.11\) is 4 sig figs. But when we multiply by \(95.211\) (5 sig figs), the limiting factor is the number of significant figures from the addition result? Wait, no. The rule is that for a series of operations, we consider the significant figures at each step.

Wait, actually, the correct approach is:

First, perform the addition: \(9.42+3.69 = 13.11\). Now, \(9.42\) has 3 significant figures, \(3.69\) has 3 significant figures. The sum \(13.11\) - when adding, the number of decimal places is determined by the least number of decimal places (2 in this case), but the number of significant figures: \(13.11\) has 4 significant figures (since all non - zero digits and zeros between non - zero digits are significant).

Now, multiply by \(95.211\) (5 significant figures). The number of significant figures in the product should be equal to the least number of significant figures in the factors. Here, the first factor (\(13.11\)) has 4 significant figures, the second factor (\(95.211\)) has 5 significant figures. So the product should have 4 significant figures.

\(13.11\times95.211 = 1248.21621\)

Now, we round \(1248.21621\) to 4 significant figures. The first four significant figures are \(1\), \(2\), \(4\), \(8\). The next digit is \(2\), which is less than 5, so we round down.

So \(1248.21621\approx1248\) (but wait, let's check again. Wait, \(13.11\times95.211\):

\(13\times95 = 1235\), \(13\times0.211 = 2.743\), \(0.11\times95 = 10.45\), \(0.11\times0.211 = 0.02321\)

Adding them up: \(1235+2.743 + 10.45+0.02321=1248.21621\)

Now, 4 significant figures: the number is \(1248.21621\). The fourth significant figure is \(8\), the next digit is \(2\), so we keep \(1248\)? Wait, no, wait \(1248\) is 4 significant figures. But wait, maybe I made a mistake in the significant figures of the sum.

Wait, \(9.42\) is 3 sig figs, \(3.69\) is 3 sig figs. When adding, \(9.42 + 3.69=13.11\). The precision is to the hundredth, but the number of sig figs: \(13.11\) has 4 sig figs. But when multiplying, the rule is that the result should have the same number of sig figs as the least precise measurement in terms of sig figs.

Wait, another way: Let's consider the original numbers. The addition is \(9.42+3.69\). \(9.42\) has 3 sig figs, \(3.69\) has 3 sig figs. The sum is \(13.11\), which we can consider as having 4 sig figs (since it's \(1.311\times10^{1}\)). Then we multiply by \(95.211\) (which is \(9.5211\times10^{1}\), 5 sig figs). When multiplying two numbers in scientific notation: \((1.311\times 10^{1})\times(9.5211\times10^{1})=(1.311\times9.5211)\times10^{2}\)

\(1.311\times9.5211 = 12.4821621\), then multiply by \(10^{2}\) gives \(1248.21621\). Now, \(1.311\) has 4 sig figs, \(9.5211\) has 5 sig figs, so the product should have 4 sig figs. So \(12.4821621\times10^{2}=1.248\times 10^{3}\) (which is \(1248\)) when rounded to 4 sig figs.

Wait, but let's check with a calculator: \(9.42 + 3.69=13.11\), \(13.11\times95.211 = 13.11\times95.211\)

\(13\times95.211 = 1237.743\)

\(0.11\times95.211 = 10.47321\)

Adding them: \(1237.743+10.47321 = 1248.21621\)

Now, to 4 significant figures: the number is \(1248.21621\). The first four significant figures are 1, 2, 4, 8. The next digit is 2, so we round down, so the result is \(1248\). But wait, maybe I made a mistake in the significant figures of the sum. Let's think again: when adding \(9.42\) (3 sig figs) and \(3.69\) (3 sig figs), the sum is \(13.11\). The uncertainty in \(9.42\) is \(\pm0.01\), and in \(3.69\) is \(\pm0.01\), so the sum has an uncertainty of \(\pm0.02\), so the sum is \(13.11\pm0.02\), which means the number of significant figures is 4 (since the last digit is uncertain, but in terms of significant figures, \(13.11\) has 4). Then when multiplying by \(95.211\) (which has an uncertainty of \(\pm0.001\)), the relative uncertainty of the sum is \(\frac{0.02}{13.11}\approx0.0015\), and the relative uncertainty of \(95.211\) is \(\frac{0.001}{95.211}\approx0.00001\). The total relative uncertainty is dominated by the sum, so the number of significant figures should be determined by the sum's significant figures (4) and the multiplier's significant figures (5), so we take 4 significant figures.

Answer:

\(1248\)