Carry out the following operation as if it were a calculation of real experimental results. Express the answer in the correct units with the correct number of significant figures. (a) 7.310 km ÷ 5.70 km(b) (3.26 X 10−3 mg) − (7.88 X 10−5 mg)(c) (4.02 X 106 dm) + (7.74 X 107 dm)(d) (7.8 m − 0.34 m)/(1.15 s + 0.82 s)(e) (9.88 × 10−3 mg) − (9.66 × 10−5 mg)

a) 1.28

b) 2.57*10^-7 mg²

c) 8.14*10^7 dm

d) 3.79 m/s

e) 9.78*10^-3 mg

Step-by-step explanation:

a) 7.31 / 5.7 = 1.28, km/km is the same unit, so they cancel each other out. The answer is 1.28 without unit.

b) (3.26*10^-3) * (7.88*10^-5) = 2.57*10^-7 mg², here, mg * mg, this is mg²

c) (4.02*10^6) + (7.74*10^7) = 8.14*10^7 dm, here, the unit are added together, so we maintain it

d) (7.8 - 0.34) m / (1.15 + 0.82) m = 7.46 m / 1.97 m = 3.79 m/s, here, the numerator has the unit of m, and the denominator has the unit of s. Giving us m/s.

e) (9.88*10^-3) - (9.66*10^-5) = 9.78*10^-3 mg. Here, the unit was subtracted from each other, so we maintain it.

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.