Question

question 3 4 pts 1) (0.7546 m)(0.89 m) = select m² 2) 143.22 cm² / 13.3 cm = select cm no new data to save. last checked at 11:57pm submit q

Explanation:

Part 1: \((0.7546\space m)(0.89\space m)\)

Step 1: Multiply the numbers

Multiply \(0.7546\) and \(0.89\).
\(0.7546\times0.89 = 0.671594\)

Step 2: Consider significant figures

\(0.89\) has two significant figures, so we round the result to two significant figures.
\(0.671594\approx0.67\) (Wait, no, wait: \(0.7546\) has four, \(0.89\) has two. When multiplying, the result should have the least number of significant figures, which is two? Wait, no, \(0.89\) is two, \(0.7546\) is four. So the product should have two significant figures? Wait, no, \(0.89\) is two, so \(0.7546\times0.89 = 0.671594\), rounded to two significant figures is \(0.67\)? Wait, no, \(0.671594\) rounded to two significant figures: the first two significant figures are 6 and 7, the next digit is 1, which is less than 5, so it's \(0.67\)? Wait, no, wait, \(0.7546\times0.89\): let's calculate again. \(0.7546\times0.89\): \(0.7546\times0.8 = 0.60368\), \(0.7546\times0.09 = 0.067914\), sum is \(0.60368 + 0.067914 = 0.671594\). Now, significant figures: \(0.89\) has two, so the result should have two. Wait, but \(0.671594\) rounded to two significant figures is \(0.67\)? Wait, no, the first significant figure is 6, second is 7, third is 1. So we keep two: \(0.67\)? Wait, but maybe the problem expects more? Wait, maybe I made a mistake. Wait, \(0.7546\) is four sig figs, \(0.89\) is two. So the product should have two sig figs. So \(0.671594\approx0.67\) (two sig figs) or maybe three? Wait, no, the rule is that when multiplying or dividing, the result has the same number of significant figures as the least precise measurement. \(0.89\) has two, so two sig figs. So \(0.67\space m^2\)? Wait, but let's check the multiplication again. Wait, \(0.7546\times0.89\): let's do it as decimals. \(0.7546\times0.89 = 0.7546\times(0.9 - 0.01) = 0.7546\times0.9 - 0.7546\times0.01 = 0.67914 - 0.007546 = 0.671594\). So, if we consider significant figures, \(0.89\) has two, so the answer should be \(0.67\space m^2\)? Wait, but maybe the problem doesn't care about significant figures and just wants the product. Then it's \(0.671594\), which can be rounded to \(0.67\) or \(0.672\) or \(0.6716\). Wait, maybe the problem expects the exact product, so \(0.671594\), which is approximately \(0.67\) (two decimal places? No, significant figures). Wait, maybe I'm overcomplicating. Let's just compute the product: \(0.7546\times0.89 = 0.671594\), so approximately \(0.67\) (or \(0.672\) if three sig figs, but \(0.89\) is two, so two).

Part 2: \(143.22\space cm^2 / 13.3\space cm\)

Step 1: Divide the numbers

Divide \(143.22\) by \(13.3\).
\(143.22\div13.3 = 10.768421\)

Step 2: Consider significant figures

\(13.3\) has three significant figures, \(143.22\) has five. So the result should have three significant figures.
\(10.768421\approx10.8\) (three significant figures: 1, 0, 7? Wait, no, 10.768... rounded to three significant figures: the first three are 1, 0, 7? Wait, no, 10.768: the first significant figure is 1, second 0, third 7, fourth 6. So we round the third: 7 + 1 = 8, so 10.8. Wait, or maybe the problem just wants the division result without significant figures, so \(10.768421\approx10.8\) (or \(10.77\) if two decimal places). Wait, let's do the division: \(143.22\div13.3\). Let's compute: \(13.3\times10 = 133\), \(143.22 - 133 = 10.22\), \(10.22\div13.3\approx0.768\), so total is \(10.768\), which is approximately \(10.8\) (three sig figs) or \(10.77\) (two decimal places).

Answer:

s:

1. \(\boldsymbol{0.67}\) (or \(0.6716\) or \(0.672\) depending on precision, but with significant figures, two sig figs give \(0.67\))
2. \(\boldsymbol{10.8}\) (or \(10.77\) depending on precision, three sig figs give \(10.8\))