Question

question
a chocolatier makes chocolate bon - bons in the shape of a sphere with a diameter of 2.6 cm. the chocolate used in the bon - bons has a density of 1.27 g/cm³. if the chocolate used costs $0.05 per gram, how much would the chocolate cost for 180 bon - bons cost, to the nearest cent?
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Explanation:

Step1: Find the radius of the sphere

The diameter of the sphere is \(2.6\) cm, so the radius \(r=\frac{2.6}{2} = 1.3\) cm.

Step2: Calculate the volume of one chocolate bon - bon (sphere)

The formula for the volume of a sphere is \(V=\frac{4}{3}\pi r^{3}\). Substitute \(r = 1.3\) into the formula:
\(V=\frac{4}{3}\pi(1.3)^{3}=\frac{4}{3}\pi\times2.197\approx\frac{4\times3.1416\times2.197}{3}\approx9.2028\space cm^{3}\)

Step3: Calculate the mass of one chocolate bon - bon

Density \( ho = 1.27\space g/cm^{3}\), and we know that \( ho=\frac{m}{V}\), so \(m= ho V\).
Substitute \( ho = 1.27\) and \(V\approx9.2028\) into the formula: \(m = 1.27\times9.2028\approx11.6876\space g\)

Step4: Calculate the cost of one chocolate bon - bon

The cost per gram is \(\$0.05\), so the cost of one bon - bon \(C_{1}=0.05\times11.6876\approx0.5844\) dollars.

Step5: Calculate the cost of 180 chocolate bon - bons

The cost of 180 bon - bons \(C = 180\times0.5844=105.192\) dollars. To the nearest cent, we look at the thousandths place. Since \(105.192\) has a thousandths digit of 2, which is less than 5, we round down. But wait, let's check the calculations again for accuracy.

Wait, let's recalculate the volume more accurately:

\(r = 1.3\), \(r^{3}=1.3\times1.3\times1.3 = 2.197\)

\(V=\frac{4}{3}\times\pi\times2.197\approx\frac{4\times3.14159265\times2.197}{3}\)

\(4\times3.14159265\times2.197 = 12.5663706\times2.197\approx27.61\)

\(V=\frac{27.61}{3}\approx9.2033\space cm^{3}\)

Mass \(m= ho V=1.27\times9.2033\approx11.6882\space g\)

Cost per bon - bon: \(0.05\times11.6882 = 0.58441\) dollars

Cost for 180 bon - bons: \(180\times0.58441 = 105.1938\) dollars, which is \(\$105.19\) when rounded to the nearest cent. But wait, maybe we made a mistake in the number of significant figures? Wait, let's do the calculation step by step with more precision.

Alternative way:

Volume of sphere: \(V=\frac{4}{3}\pi r^{3}\), \(r = 1.3\)

\(V=\frac{4}{3}\times3.1416\times1.3^{3}=\frac{4}{3}\times3.1416\times2.197\)

\(4\times3.1416 = 12.5664\)

\(12.5664\times2.197 = 12.5664\times2+12.5664\times0.197=25.1328 + 2.4755808 = 27.6083808\)

\(V=\frac{27.6083808}{3}=9.2027936\space cm^{3}\)

Mass \(m = 1.27\times9.2027936=1.27\times9 + 1.27\times0.2027936=11.43+0.257547872 = 11.687547872\space g\)

Cost per bon - bon: \(0.05\times11.687547872 = 0.5843773936\) dollars

Cost for 180 bon - bons: \(180\times0.5843773936 = 105.1879308\) dollars, which is approximately \(\$105.19\) when rounded to the nearest cent. Wait, but maybe we should check if we need to round during each step or at the end. If we round at the end, \(105.19\) dollars. But let's see, maybe there is a miscalculation. Wait, let's check the density and the cost again.

Wait, the cost per gram is \(\$0.05\), which is 5 cents per gram.

Wait, another way:

First, volume of one bon - bon:

\(r=\frac{2.6}{2}=1.3\) cm

\(V=\frac{4}{3}\pi r^{3}=\frac{4}{3}\times3.1416\times1.3^{3}\)

\(1.3^{3}=2.197\)

\(\frac{4}{3}\times3.1416\times2.197 = \frac{12.5664\times2.197}{3}\)

\(12.5664\times2.197 = 27.608\) (approx)

\(V = 9.2027\space cm^{3}\)

Mass \(m= ho V=1.27\times9.2027 = 11.6874\space g\)

Cost per bon - bon: \(0.05\times11.6874 = 0.58437\) dollars

Cost for 180: \(180\times0.58437 = 105.1866\) dollars, which is \(\$105.19\) when rounded to the nearest cent.

Wait, but maybe the problem expects us to use \(\pi = 3.14\) for simplicity. Let's try that:

\(V=\frac{4}{3}\times3.14\times1.3^{3}=\frac{4}{3}\times3.14\times2.197=\frac{12.56\times2.197}{3}=\frac{27.69432}{3}=9.23144\space cm^{3}\)

Mass \(m = 1.27\times9.23144 = 11.7239\space g\)

Cost per bon - bon: \(0.05\times11.7239 = 0.586195\) dollars

Cost for 180: \(180\times0.586195 = 105.5151\) dollars, which is \(\$105.52\)? Wait, this is a big difference. Wait, maybe I made a mistake in the radius? No, radius is half of diameter, diameter is 2.6, so radius is 1.3. Wait, maybe the problem is in the density or the cost? Wait, the problem says "to the nearest cent", so let's see.

Wait, let's do the calculation with more precise steps:

1. Radius \(r=\frac{2.6}{2}=1.3\) cm.

2. Volume of sphere: \(V = \frac{4}{3}\pi r^{3}=\frac{4}{3}\times\pi\times(1.3)^{3}\)

\((1.3)^{3}=1.3\times1.3\times1.3 = 2.197\)

\(V=\frac{4}{3}\times\pi\times2.197\approx9.2028\space cm^{3}\)

3. Mass of one bon - bon: \(m= ho V = 1.27\times9.2028\approx11.6876\space g\)

4. Cost of one bon - bon: \(C_{1}=0.05\times11.6876 = 0.58438\) dollars

5. Cost of 180 bon - bons: \(C = 180\times0.58438=105.1884\) dollars, which is approximately \(\$105.19\) when rounded to the nearest cent.

Wait, but maybe the answer is \(\$105.19\) or maybe I made a mistake. Wait, let's check the multiplication again. \(180\times0.58438 = 105.1884\), which is \(\$105.19\) when rounded to the nearest cent.

Answer:

\(\$105.19\)