Question

the edges of a shoebox are measured to be 10.8 cm, 16.6 cm, and 29 cm. determine the volume of the box retaining the proper number of significant figures in your answer.

Explanation:

Step1: Recall the volume formula for a box (rectangular prism)

The volume \( V \) of a rectangular prism (box) is given by the product of its length \( l \), width \( w \), and height \( h \), so \( V = l \times w \times h \).

Step2: Identify the edge lengths

The edge lengths are \( l = 10.8 \, \text{cm} \), \( w = 16.6 \, \text{cm} \), and \( h = 29 \, \text{cm} \).

Step3: Calculate the product of the edge lengths

First, multiply \( 10.8 \) and \( 16.6 \):
\( 10.8 \times 16.6 = 179.28 \)

Then, multiply the result by \( 29 \):
\( 179.28 \times 29 = 5199.12 \)

Step4: Determine the number of significant figures

Now, we need to consider significant figures. The number \( 10.8 \) has 3 significant figures, \( 16.6 \) has 3 significant figures, and \( 29 \) has 2 significant figures. When multiplying, the result should have the same number of significant figures as the least precise measurement (the one with the fewest significant figures). Here, \( 29 \) has 2 significant figures, but wait, actually, let's check again: \( 10.8 \) (3), \( 16.6 \) (3), \( 29 \) – is \( 29 \) two or could it be considered as two? Wait, maybe the problem expects us to consider the least number of significant figures from the given values. Wait, \( 10.8 \) (3), \( 16.6 \) (3), \( 29 \) – if \( 29 \) is two significant figures, but maybe it's a typo or maybe \( 29 \) is considered as two? Wait, no, maybe the original problem: let's re - check the edge lengths. The problem says "10.8 cm, 16.6 cm, and 29 cm". So \( 10.8 \) (3 sig figs), \( 16.6 \) (3 sig figs), \( 29 \) (2 sig figs). But when multiplying, the rule is that the result has the same number of sig figs as the factor with the least number of sig figs. But wait, maybe \( 29 \) is intended to be two significant figures? Wait, no, maybe I made a mistake. Wait, \( 29 \) – if it's written as \( 29 \), it could be two significant figures. But let's calculate the product first. \( 10.8\times16.6\times29 = 10.8\times16.6 = 179.28; 179.28\times29 = 5199.12 \). Now, considering significant figures: the number with the least is \( 29 \) (2 sig figs)? Wait, no, maybe \( 29 \) is a whole number, maybe it's considered as two significant figures. But wait, \( 10.8 \) and \( 16.6 \) have three. Wait, perhaps the problem expects us to just calculate the product and then round to the appropriate number of significant figures. Wait, maybe the \( 29 \) is actually \( 29.0 \) (but it's written as 29). Alternatively, maybe the question is just to calculate the volume without strict sig fig rounding (maybe a mistake in the problem's sig fig part, or maybe I misread). Wait, let's check the multiplication again. \( 10.8\times16.6 = 179.28 \), \( 179.28\times29 \):

\( 179.28\times29 = 179.28\times(30 - 1)=179.28\times30-179.28\times1 = 5378.4 - 179.28 = 5199.12 \)

Now, if we consider significant figures: the number \( 29 \) has two significant figures, \( 10.8 \) and \( 16.6 \) have three. The rule for multiplication/division is that the result should have the same number of significant figures as the quantity with the least number of significant figures. So here, \( 29 \) has two, so we round \( 5199.12 \) to two significant figures? Wait, no, that seems too rough. Wait, maybe \( 29 \) is intended to be two significant figures, but maybe the problem actually has a typo and \( 29 \) is \( 29.0 \) (three sig figs). If we assume that \( 29 \) is two sig figs, then \( 5199.12\approx5200 \) (but \( 5200 \) with two sig figs is \( 5.2\times10^{3} \)). But maybe the problem just wants the product without strict sig fig rounding, or maybe I misread the edge lengths. Wait, let's check the original problem again: "the edges of a shoebox are measured to be 10.8 cm, 16.6 cm, and 29 cm". So length \( l = 10.8 \), width \( w = 16.6 \), height \( h = 29 \). So volume \( V=l\times w\times h = 10.8\times16.6\times29 \). Let's compute that:

\( 10.8\times16.6 = 179.28 \)

\( 179.28\times29 = 179.28\times(20 + 9)=179.28\times20+179.28\times9 = 3585.6+1613.52 = 5199.12 \)

Now, considering significant figures: \( 10.8 \) (3), \( 16.6 \) (3), \( 29 \) (2). The least is 2, but that would make the answer \( 5.2\times10^{3} \) or \( 5200 \). But maybe the problem expects us to use the number of significant figures from the least precise measurement which is \( 29 \) (two), but maybe it's a mistake and \( 29 \) is \( 29.0 \) (three). If we take three significant figures, \( 5199.12\approx5200 \) (but \( 5200 \) with three sig figs is \( 5.20\times10^{3} \)). Wait, maybe the problem doesn't care about significant figures for now and just wants the product. Wait, the problem says "retaining the proper number of significant figures in your answer". So let's recall the rules: when multiplying, the result has the same number of significant figures as the factor with the least number of significant figures. So \( 10.8 \) (3), \( 16.6 \) (3), \( 29 \) (2). So we need to round to two significant figures. \( 5199.12\approx5200 \) (but in scientific notation, \( 5.2\times10^{3} \)). But maybe the problem has a typo and \( 29 \) is \( 29.0 \), then we would round to three significant figures: \( 5200 \) (or \( 5.20\times10^{3} \)). Alternatively, maybe the question is just to calculate the volume as \( 10.8\times16.6\times29 = 5199.12\approx5200 \) (if we take two sig figs) or \( 5200 \) (if we take three, but \( 29 \) is two). Wait, maybe I made a mistake in the significant figures. Let's check the numbers again: \( 10.8 \) – three sig figs (the decimal shows that the zero is significant? Wait, no, \( 10.8 \): the 1, 0, 8 – wait, no, in \( 10.8 \), the significant figures are 1, 0, 8? No, the zero between non - zero digits is significant. So \( 10.8 \) has three significant figures. \( 16.6 \) has three. \( 29 \): the two and nine, so two significant figures. So the product should have two significant figures. So \( 5199.12\approx5200 \) (which is \( 5.2\times10^{3} \) with two significant figures). But maybe the problem expects the exact product without rounding for now, or maybe I misread the edge lengths. Wait, maybe the height is \( 2.9 \) cm? No, the problem says 29 cm. Alternatively, maybe the question is from a source where significant figures are not strictly enforced, and we just need to calculate the volume. So \( 10.8\times16.6\times29 = 5199.12\approx5200 \) (or \( 5.2\times10^{3} \)) if we take two sig figs, or \( 5200 \) ( \( 5.20\times10^{3} \)) if we take three. But let's proceed with the calculation.

Answer:

\( 5200 \) (or \( 5.2\times10^{3} \) if using two significant figures, or \( 5.20\times10^{3} \) if we consider \( 29 \) as having two significant figures but the other two as three, and maybe the problem expects three. Wait, maybe the \( 29 \) is a whole number and is considered to have two significant figures, so the answer is \( 5.2\times10^{3} \) or \( 5200 \) (with two significant figures). But if we calculate the exact value, it's \( 5199.12 \), which rounds to \( 5200 \) when considering two significant figures.