Let's attack this problem using the z-score concept. The sample std. dev. here is (0.25 oz)/sqrt(40), or 0.040. Thus, the z score representing 3.9 oz. is
3.9 - 4.0
z = -------------- = -2.5
0.040
In one way or another we must find the area under the std. normal curve that lies to the left of z = -2.5. Use a table of z-scores or a calculator with built-in statistics functions. According to my TI-83 Plus calculator, that area is
0.006. One way of interpreting this that with so small a standard deviation, most volumes of coffee put into the jars are very close to the mean, 4 oz.
domain of the problem is -5 and the range is 4
So 2 gallons every 5 minutes
2/5= .4
so .4 a minute
and you already have 5 gallons in so those need to be added
m=minutes
y=0.4m + 5 will be what you want to find out if you are looking to find out how much will be there in a certain time
for 50 minutes you will have
y=0.4(50) +5
20+5
25 gallons
HOWEVER, the equation has to be changed if you want to tell how long you have to wait for it to fill.
m=2.5(g-5)
m=minutes
g= gallons
you subtract 5 because they are already there
you multiply by 2.5 because it fills at a rate of 1 gallon every 2.5 minutes.
m=2.5(1500-5)
for the sake of it being easier i will do the -5 separately
2.5(1500 = 3750
2.5(-5= -12.5
3750-12.5
3737.5 minutes to fill the pool.
3720/60 = 62
17.5/60 = .292
62.292 hours to fill the pool.
p.s. you have a really slow hose.
Its good you didn't wait for it to fill, you would have died from lack of water before then if you just sat and waited.
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Answer:
Step-by-step explanation:
It was 12 take away 210 add 20 to 3243
