Question

a hot air balloon holds 4,328 cubic meters of helium. the density of helium is 0.1785 kilograms per cubic meter. how many kilograms of helium does the balloon contain, rounded to the nearest tenth of a kilogram? 3,747.9 kg 3,875.3 kg 772.5 kg 4,383.2 kg

Explanation:

Step1: Use the density formula

The formula for density is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass, and $V$ is volume. We want to find $m$, so we can rewrite the formula as $m =
ho\times V$.

Step2: Substitute the given values

We are given that $V = 4328$ cubic meters and $
ho=0.1785$ kilograms per cubic meter. Substituting these values into the formula $m=
ho\times V$, we get $m = 0.1785\times4328$.

Step3: Calculate the product

$0.1785\times4328=(0.1 + 0.07+0.008 + 0.0005)\times4328=0.1\times4328+0.07\times4328+0.008\times4328+0.0005\times4328=432.8+302.96+34.624 + 2.164=772.548$.

Step4: Round to the nearest tenth

Rounding $772.548$ to the nearest tenth. Looking at the digit in the hundredth place, which is $4$. Since $4<5$, we keep the digit in the tenth place as it is. So $772.548\approx772.5$.

Answer:

$772.5$ kg