A hose is left running for 240 minutes to 2 significant figures. The amount of water coming out of the hose each minute is 2.1 litres to 2 significant figures. Calculate the lower and upper bounds of the total amount of water that comes out of the hose.​

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caylus

Answer:

Hello,

Step-by-step explanation:

Let say t the time the hose is left running

235 ≤ t < 245 (in min)

Let say d the amount of water coming out of the hose each minute

2.05 ≤ d < 2.15 (why d : débit in french)

235*2.05 ≤ t*d < 245*2.15

481.75 ≤ t*d < 526.75    (litres)

Answer:

Lower bound: [tex]495\; \rm L[/tex] (inclusive.)

Upper bound: [tex]505\; \rm L[/tex] (exclusive.)

Step-by-step explanation:

The amount of water from the hose is the product of time and the rate at which water comes out.

When multiplying two numbers, the product would have as many significant figures as the less accurate factor.

In this example, both factors are accurate to two significant figures. Hence, the product would also be accurate to two significant figures. That is:

[tex]240 \times 2.1 = 5.0 \times 10^{2}\; \rm L[/tex] ([tex]500\; \rm L[/tex] with only two significant figures.)

Let [tex]x[/tex] denote the amount of water in liters. For [tex]x\![/tex] to round to [tex]5.0 \times 10^{2}\; \rm L[/tex] only two significant figures are kept, [tex]495 \le x < 505[/tex]. That gives a bound on the quantity of water from the hose.