Question

Which table has a constant of proportionality between y and x of 1/6?

choose 1 answer.

A: x--15 19&1/2 36
y---5 6 & 1/2 12

B: x--12 13&1/2 24
y---2 3 &1/2 14

C: x--18 27 23
y--3 4&1/2 5&1/2

Plz help this is just way to confusing. :>

Answer 1

C is the right answer.

Explanation:

Khan Academy

Answer 2

Your Question is not well presented.

See Question Below

Which table has a constant of proportionality between y and x of 1/6? (choose 1 answer.)

A:

x--> 15 ---19
`(1)/(2)` --- 36

y--> 5 ---- 6
`(1)/(2)` ---- 12

B:

x--> 12 ---13
`(1)/(2)` --- 24

y--> 2 ---- 3
`(1)/(2)` ---- 14

C:

x--> 18 --- 27 --- 33

y--> 3 ---- 4
`(1)/(2)` ---- 5
`(1)/(2)`

Answer:

Table C has 1/6 as the constant of proportionality between y and x

Explanation:

Given

Table A, B, C

Required

To check which of the tables has a constant of proportionality of 1/6

The constant of proportionality is calculated by dividing individual values of y column with x column.

Mathematically, this is represented by

`k = (y)/(x)`

Where k is the constant of proportionality

Recall Table A

x--> 15 ---19
`(1)/(2)` --- 36

y--> 5 ---- 6
`(1)/(2)` ---- 12

When x = 15, y = 5.

The constant of proportionality becomes

`k = (y)/(x)`

`k = (5)/(15)` --- Simplify fraction to lowest term by dividing by 5

`k = (1)/(3)`

So, when x = 15, y = 5.

`k = (1)/(3)`

`(1)/(3)` is not equal to
`(1)/(6)`; So, we do not need to check further in table A.

Hence, table A does not have 1/6 as the constant of proportionality between y and x

We move to table B

Recall Table B

x--> 12 ---13
`(1)/(2)` --- 24

y--> 2 ---- 3
`(1)/(2)` ---- 14

When x = 12, y = 2.

The constant of proportionality becomes

`k = (y)/(x)`

`k = (2)/(12)` --- Simplify fraction to lowest term by dividing by 2

`k = (1)/(6)`

We can't conclude yet, if the constant of proportionality between y and x in table B is
`(1)/(6)` until we check further

When
`x = 3(1)/(2) , y = 13(1)/(2)`

The constant of proportionality becomes

`k = (y)/(x)`

`k = (3(1)/(2))/(13(1)/(2))` --- Convert to decimal

`k = (3.5)/(13.5)` Simplify fraction to lowest term by dividing by 0.5

`k = (3.5/0.5)/(13.5/0.5)`

`k = (7)/(27)` -- This cannot be simplified any further

`(7)/(27)` is not equal to
`(1)/(6)`; So, we do not need to check further in table B.

Hence, table B does not have 1/6 as the constant of proportionality between y and x

We move to table C

Recall Table C

x--> 18 --- 27 --- 33

y--> 3 ---- 4
`(1)/(2)` ---- 5
`(1)/(2)`

When x = 18, y = 3

The constant of proportionality becomes

`k = (y)/(x)`

`k = (3)/(18)` --- Simplify fraction to lowest term by dividing by 3

`k = (1)/(6)`

We can't conclude yet, if the constant of proportionality between y and x in table C is
`(1)/(6)` until we check further

When x = 27,
`y = 4(1)/(2)`

The constant of proportionality becomes

`k = (y)/(x)`

`k = (4(1)/(2))/(27)` --- Convert to fraction to decimal

`k = (4.5)/(27)` Simplify fraction to lowest term by dividing by 4.5

`k = (1)/(6)`

We still can't conclude until we check further

When x = 33,
`y = 5(1)/(2)`

The constant of proportionality becomes

`k = (y)/(x)`

`k = (5(1)/(2))/(33)` --- Convert to fraction to decimal

`k = (5.5)/(33)` Simplify fraction to lowest term by dividing by 5.5

`k = (1)/(6)`

Notice that; for every value of x and its corresponding value of y, the constant of proportionality, k maintains
`(1)/(6)` as its value

Hence, we can conclude that "Table C has 1/6 as the constant of proportionality between y and x"