The table shows a proportional relationship between the mass, in kilograms
(
kg
)
(kg)left parenthesis, start text, k, g, end text, right parenthesis, of a dog and the milliliters
(
mL
)
(mL)left parenthesis, start text, m, L, end text, right parenthesis of flea medicine a veterinarian prescribes.
Flea medicine prescriptions
Dog's mass
(
kg
)
(kg)left parenthesis, start text, k, g, end text, right parenthesis Medicine
(
mL
)
(mL)left parenthesis, start text, m, L, end text, right parenthesis
50
5050
2
22
10
1010
0.4
0.40, point, 4
5
55
0.2
0.20, point, 2
? ?
A row of values is missing in the table.
Which of the following numbers of kilograms and milliliters could be used as the missing values in the table?
Choose 2 answers:
Choose 2 answers:

(Choice A)
A
Dog:
15

kg
15kg15, start text, k, g, end text
Medicine:
0.9

mL
0.9mL0, point, 9, start text, m, L, end text

(Choice B)
B
Dog:
25

kg
25kg25, start text, k, g, end text
Medicine:
0.8

mL
0.8mL0, point, 8, start text, m, L, end text

(Choice C)
C
Dog:
40

kg
40kg40, start text, k, g, end text
Medicine:
1.6

mL
1.6mL1, point, 6, start text, m, L, end text

(Choice D)
D
Dog:
35

kg
35kg35, start text, k, g, end text
Medicine:
1.4

mL
1.4mL1, point, 4, start text, m, L, end text

(Choice E)
E
Dog:
20

kg
20kg20, start text, k, g, end text
Medicine:
1.2

mL
1.2mL1, point, 2, start text, m, L, end text

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form [tex]k=\frac{y}{x}[/tex] or [tex]y=kx[/tex]

Let

x ---> the mass, in kilograms of a dog

y ---> milliliters of flea medicine a veterinarian prescribes

Find the value of the constant of proportionality k for the given data

[tex]k=\frac{y}{x}[/tex]

For x=50 kg, y=2 mL ----> [tex]k=\frac{2}{50}=0.04\ mL/kg[/tex]

For x=10 kg, y=0.4 mL ----> [tex]k=\frac{0.4}{10}=0.04\ mL/kg[/tex]

For x=5 kg, y=0.2 mL ----> [tex]k=\frac{0.2}{5}=0.04\ mL/kg[/tex]

so

The linear direct equation is equal to

[tex]y=0.04x[/tex]

Verify each choice

A) Dog: 15 kg, Medicine: 0.9 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

[tex]\frac{0.9}{15}= 0.06\ mL/kg[/tex]

so

[tex]0.06 \neq 0.04[/tex]

therefore

These numbers could not be used as the missing values in the table

B) Dog: 25 kg, Medicine: 0.8 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

[tex]\frac{0.8}{25}= 0.032\ mL/kg[/tex]

so

[tex]0.032 \neq 0.04[/tex]

therefore

These numbers could not be used as the missing values in the table

C) Dog: 40 kg, Medicine: 1.6 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

[tex]\frac{1.6}{40}= 0.04\ mL/kg[/tex]

so

[tex]0.04 = 0.04[/tex]

therefore

These numbers could be used as the missing values in the table

D) Dog: 35 kg, Medicine: 1.4 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

[tex]\frac{1.4}{35}= 0.04\ mL/kg[/tex]

so

[tex]0.04 \neq 0.04[/tex]

therefore

These numbers could be used as the missing values in the table

E) Dog: 20 kg, Medicine: 1.2 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

[tex]\frac{1.2}{20}= 0.06\ mL/kg[/tex]

so

[tex]0.06 \neq 0.04[/tex]

therefore

These numbers could not be used as the missing values in the table

ninjathan300 ninjathan300 · Answer 2 · 2020-12-10T18:33:06+01:00

ninjathan300 ninjathan300

10-12-2020

Answer:

Dog: 40 kg

Medicine: 1.6mL

Dog: 35 kg

Medicine: 1.4mL

Step-by-step explanation:

This combination has 0.040.040, point, 04 milliliters of medicine per kilogram of the dog's mass, which is the same as in the table.

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