Uche pumps gasoline at a rate of 18\,\dfrac{\text{L}}{\text{min}}18 min L 18, start fraction, start text, L, end text, divided by, start text, m, i, n, end text, end fraction. What is Uche's pumping rate in \dfrac{\text{mL}}{\text{s}} s mL start fraction, start text, m, L, end text, divided by, start text, s, end text, end fraction?

Uche's pumping rate is 300 mL/s.

Step-by-step explanation:

Given:

Uche pumps gasoline at a rate of 18 L/min.

Now, to find Uche's pumping rate in mL/s.

The rate at which Uche pumps gasoline = 18 L/min.

So, to get the pumping rate in mL/s we use convert L/min to mL/s by using conversion factor:

1 L/min = 16.6667 mL/s.

18 L/min = 16.6667 × 18 mL/s.

18L/min = 300 mL/s.

Therefore, Uche's pumping rate is 300 mL/s.

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

step-by-step explanation:

the answer is 89 d

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