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Verdich [7]
2 years ago
8

A movie theater sells popcorn in bags of different sizes. The table shows the volume of popcorn and the price of the bag.

Mathematics
1 answer:
scoundrel [369]2 years ago
6 0

Answer:

<em>The price for a 60-ounce bag of popcorn would be $16</em>

Step-by-step explanation:

<u>Function Modeling</u>

The behavior of some parameters that depend on a set of variables can be modeled in several ways, like linear, quadratic, exponential, logarithmic, among many others.

The selection of the model is often a complex decision that involves statistics and data analysis.

The question provides us with four points where the volume of popcorn bags and the price in dollars. The easiest function that can be used is the line.

The equation of a line of the volume V and the price p can be found with the expression

\displaystyle p-p_1=\frac{p_2-p_1}{V_2-V_1}(V-V_1)

We'll use the first two values (6,10) (8,20)

\displaystyle p-6=\frac{8-6}{20-10}(V-10)

Simplifying and rearranging, we get the model

\displaystyle p(V)=\frac{1}{5}V+4

To test the accuracy of the model, we compute the values of p for V=35 and for V=48

\displaystyle p(35)=\frac{1}{5}(35)+4=11

\displaystyle p(48)=\frac{1}{5}(48)+4=13.6

Since the computed values are equal to those of the table, the model is accurate. We can now predict the price for V=60

\displaystyle p(60)=\frac{1}{5}(60)+4=16

The price for a 60-ounce bag of popcorn would be $16

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<u>EXPLANATION</u>

We want to find the cube root of

- 729 {a}^{9}  {b}^{6}

We express this symbolically as:

\sqrt[3]{- 729 {a}^{9}  {b}^{6} }

The expression under the radical called the radicand.

We need to express this radical in exponential form using the property,

{x}^{ \frac{m}{n} }  =  \sqrt[n]{ {x}^{m} }

Applying this rule gives us:

\sqrt[3]{- 729 {a}^{9}  {b}^{6} }  =  ({- 729 {a}^{9}  {b}^{6}})^{ \frac{1}{3} }

\sqrt[3]{- 729{a}^{9}  {b}^{6} }  =  ({- {9}^{3}  {a}^{9}  {b}^{6}})^{ \frac{1}{3} }

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We apply this rule on the RHS to get,

\sqrt[3]{- 729{a}^{9}  {b}^{6} }  =  ({- {9}^{3 \times { \frac{1}{3} } }  {a}^{9 \times { \frac{1}{3} } }  {b}^{6 \times { \frac{1}{3} } }})

This simplifies to

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Step 7 given

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