Chapter 1.2, Problem 36PPE

(a)

To determine

To find: the volume of to nearest tenth of cubic inches juice can at the right.

Answer to Problem 36PPE

  $V=23.96448{\text{m}}^3$ .

Explanation of Solution

Given:

  

Concept used:

The shape of can is cylindrical and the volume of the cylinder is:

  $\begin{array}{l}V=\pi r^{2}h.\\ \pi =\mathrm{3.14.}\end{array}$

Calculation:

Consider the following formula for volume of a cylinder:

  $V=\pi r^{2}h\mathrm{......}\left(i\right)$

Here, $r$ is the radius and $h$ is the height of given cylinder.

  $\left(a\right)$

Now, substitute $\pi =3.14,r=1.2\text{ and }h=5.3$ in the expression $\left(i\right)$ :

  $V=3.14\times {\left(1.2\right)}^2\times 5.3$ .

On simplifying the power, it becomes:

  $V=3.14\times 1.44\times 5.3$ .

Finally, on multiplying all the terms, it becomes:

  $V=23.96448{\text{m}}^3$ .

Hence, $V=23.96448{\text{m}}^3$ .

(b)

To determine

To find: the reason to the number of can at the nearest tenth of cubic inches and does a fluid ounce of juice fill.

Answer to Problem 36PPE

The nearest tenth place rounding off the amount of juice filled in the cylinder is:

  $6.6{\text{in}}^3$ .

Explanation of Solution

Given:

  

Concept used:

Conversion of fluid ounce to cubic inches or vice versa:

  $1floz=0.55411{\text{in}}^3$ .

The shape of can is cylindrical and the volume of the cylinder is:

  $\begin{array}{l}V=\pi r^{2}h.\\ \pi =\mathrm{3.14.}\end{array}$

Calculation:

Now, for rounding to nearest tenth, first locate the digit in the tenth place of given decimal number, that is:

  $23.96448{\text{m}}^3$ .

Clearly, the digit in the tenth place of given decimal number is $9$ .

Further, it checks the next digit to the given located digit.

Clearly, the digit next to $9\text{ is }6$ .

Now, applying the rule of round off, since the digit is $6$ , greater than $5$ .

So, the digit $9$ is rounded up to $0$ and the digit at ones place that is $3$ and rounded up to $4$ .

Thus, the required rounding off to nearest tenth place of the volume $\left(V\right)$ for the cylinder, that is:

  $23.96448{\text{in}}^3$ is as follows:

Hence, the nearest tenth place rounding off of the volume for the cylinder is $24.0{\text{in}}^3$ .

Next is to find the amount of juice filled in the cylinder to the nearest tenth of the cubic inches. Consider the amount of juice filled in fluid ounce is $12floz$ .

As,

  $1floz=0.55411{\text{in}}^3$

The amount of juice filled in cubic inches:

  $\begin{array}{l}\left(12\times 0.55411\right){\text{in}}^3.\\ \Rightarrow 6.64932{\text{in}}^3.\end{array}$

Now, for rounding to nearest tenth, first locate the digit in the tenth place of given decimal number that is:

  $6.64932{\text{in}}^3$ .

Clearly, the digit in the tenth place of given decimal number, that is $6.64932{\text{in}}^3\text{ is }6$ .

Further, it checks the next digit to the given located digit.

Clearly, the digit next to $6\text{ is }4$ .

Now, applying the rule of round off, since, the digit is $4$ , less than $5$ .

So, the digit $6$ is rounded down to $6$ .

Thus, the required rounding off to nearest tenth place of the amount of juice filled in the cylinder.

That is, $6.64932{\text{in}}^3$ is as follows:

Hence, the nearest tenth place rounding off the amount of juice filled in the cylinder is:

  $6.6{\text{in}}^3$ .