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2 years ago

Let fat be the explanatory variable and calories be the response variable; which scenario best describes ŷ = 121 + 1.42x?

Mathematics

1 answer:

emmasim [6.3K]2 years ago

8 0

Using linear function concepts, it is found that the correct option is:

The amount of calories in a cheeseburger increases by 1.42 for every one gram of fat. The calorie amount estimated by this model is 121 if there are zero grams of fat.

A linear function is modeled by:

$y = mx + b$

In which:

m is the slope, which represents by how much y changes when x changes by 1.
b is the y-intercept, which is the value of y when x = 0.

In this problem:

The explanatory variable fat is the input.
The response variable calories is the output.

The equation is:

$y = 121 + 1.42x$

The slope is of 1.42, which means that the amount of calories increases by 1.42 when the amount of fat increases by 1 gram.
The y-intercept is of 121, which means that if there are 0 grams of fat, there will be 121 calories.

Thus, the correct option is:

The amount of calories in a cheeseburger increases by 1.42 for every one gram of fat. The calorie amount estimated by this model is 121 if there are zero grams of fat.

A similar problem is given at brainly.com/question/16302622

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COMPUTE 3 ( 2 1/2 - 1 ) + 3/10

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Answer:

<h3> $\boxed{ \frac{24}{5} }$ </h3>

Step-by-step explanation:

$\mathsf{3(2 \frac{1}{2} - 1) + \frac{3}{10} }$

Convert mixed number to improper fraction

$\mathrm{3( \frac{5}{2} - 1) + \frac{3}{10} }$

Calculate the difference

⇒ $\mathrm{3( \frac{5 \times 1}{2 \times 1} - \frac{1 \times 2}{1 \times 2} }) + \frac{3}{10}$

⇒ $\mathrm{ 3 \times( \frac{5}{2} - \frac{2}{2}) } + \frac{3}{10}$

⇒ $\mathrm{3 \times ( \frac{5 - 2}{2} ) + \frac{3}{10} }$

⇒ $\mathrm{3 \times \frac{3}{2} + \frac{3}{10} }$

Calculate the product

⇒ $\mathrm{ \frac{3 \times 3}{1 \times 2} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{9}{2} + \frac{3}{10}}$

Add the fractions

⇒ $\mathsf{ \frac{9 \times 5}{2 \times 5} + \frac{3 \times 1}{10 \times 1} }$

⇒ $\mathrm{ \frac{45}{10} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{45 + 3}{10 } }$

⇒ $\mathrm{ \frac{48}{10} }$

Reduce the numerator and denominator by 2

⇒ $\mathrm{ \frac{24}{5} }$

Further more explanation:

Addition and Subtraction of like fractions

While performing the addition and subtraction of like fractions, you just have to add or subtract the numerator respectively in which the denominator is retained same.

For example :

Add : $\mathsf{ \frac{1}{5} + \frac{3}{5} = \frac{1 + 3}{5} } = \frac{4}{5}$

Subtract : $\mathsf{ \frac{5}{7} - \frac{4}{7} = \frac{5 - 4}{7} = \frac{3}{7} }$

So, sum of like fractions : $\mathsf{ = \frac{sum \: of \: their \: number}{common \: denominator} }$

Difference of like fractions : $\mathsf{ \frac{difference \: of \: their \: numerator}{common \: denominator} }$

Addition and subtraction of unlike fractions

While performing the addition and subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fractions. Thus, obtained fractions should be reduced into lowest terms if there are any common on numerator and denominator.

For example:

$\mathsf{add \: \frac{1}{2} \: and \: \frac{1}{3} }$

L.C.M of 2 and 3 = 6

So, ⇒ $\mathsf{ \frac{1 \times 3}{2 \times 3} + \frac{1 \times 2}{3 \times 2} }$

⇒ $\mathsf{ \frac{3}{6} + \frac{2}{6} }$

⇒ $\frac{5}{6}$

Multiplication of fractions

To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term.

When any number or fraction is divided by a fraction, we multiply the dividend by reciprocal of the divisor. Let's consider a multiplication of a whole number by a fraction:

$\mathsf{4 \times \frac{3}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6}$

Multiplication for $\mathsf{ \frac{6}{5} \: and \: \frac{25}{3} }$ is done by the similar process

$\mathsf{ = \frac{6}{5} \times \frac{25}{3} = 2 \times 5 \times 10}$

Hope I helped!

Best regards!

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