Q6. - Algebra 1

Question

The table shows the relationship between calories and fat grams contained in orders of French fries from various restaurants.

Calories	Fat Grams
240	14
280	15
310	16
260	12
340	16
350	18
300	13

Assuming the data can best be described by a linear model, how many fat grams would be expected to be contained in a 315-calorie order of French fries?

 

F 15 fat grams

G 16 fat grams

H 17 fat grams

J 18 fat grams

Step-by-Step Solution

Answer

Option F is the correct answer.

Step 1. Define a variable.

A variable in Mathematics is defined as the alphabetic character that expresses a numerical value or a number.

Step 2. Define an equation.

In mathematics, an equation is a statement that asserts the equality of two expressions, which are connected by the sign $"="$.

Step 3. Complete the number of fat grams to be expected in a 315-calorie order of French fries.

Assume the data can best be described by a linear model.

Let the linear equation is 

 $y=ax+b\mathrm{........}\left(1\right)$.

Where, 

$y=$ The number of fat grams

$x=$ The number of calories

$a$ and $b$ are the constants.

Observe the table.

At $x=260,y=12$

Use equation (1).

$12=260a+b\mathrm{........}\left(2\right)$

At $x=340,y=16$

Use equation (1).

$16=340a+b\mathrm{.........}\left(3\right)$

Subtract (2) from (3).

$\begin{array}{c}4=80a\\ a=\frac{4}{80}\\ a=\frac{1}{20}\end{array}$

Substitute $a=\frac{1}{20}$ in equation (2).

$$\begin{gathered} 12=260\left(\frac{1}{20}\right)+b \\ 12=13+b \\ b=12-13 \\ b=-1 \end{gathered}$$

Substitute $a=\frac{1}{20}$ and $b=-1$, in equation (1).

$$\begin{gathered} y=\frac{1}{20}x+\left(-1\right) \\ y=\frac{1}{20}x-1\mathrm{........}\left(4\right) \end{gathered}$$

Now, substitute $x=315$ in equation (4).

$$\begin{gathered} y=\frac{1}{20}\left(315\right)-1 \\ y=15.75-1 \\ y=14.75 \\ y\approx 15 \end{gathered}$$ 

Hence, 15 fat grams would be expected in a 315-calorie order of French fries.

 

Therefore, option F is the correct answer.