What you don't want is the value of r(t) becoming negative. Surely that would represent water escaping the reservoir.
How big can (t) get before water actually starts escaping the reservoir?
Essentially, to figure this out r(t) would have to be equal to 0.
700 - 40t = 0
40t=700
t=700/40=17.5
So the first answer is 17.5 seconds. After this amount of time has elapsed the reservoir will start to lose water as r(t) would become negative.
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The reservoir had the least amount of water in it before it was being filled. That was when t=0. The volume of water in the reservoir wasn't negatively impacted as not enough water had escaped it during the 17.5 to 30 second period.
2) 1/16
3) 25/3 or 8 and 1/3
4) to find the reciprocal of 4 we must make 4 a fraction. Simply add a one as the denominator and then switch as you would usually do when finding the reciprocal.
4 = 4/1 = 1/4
Answer : 1/4
5) first, let's convert 4 5/8 into a improper fraction.
4 5/8 = 37/8
Now let's find the reciprocal of that number...
Answer : 8/37
6) 5/8 ÷ 2/3 does equal 15/16
To check, well do the inverse operation.
The opposite of division is multiplication.
Answer : d
7) 6
8) 13
It would go like this 6(b-4)>30 would then be 6b-24>30 then you would have to add 24 to both the -24 and the 30 to get 6b>54 the you would have to divide 54 by 6 and after that your answer would be b> 9
<h3>
Answer: Approximately 2.33</h3>
The variance is given to be 5.43
Apply the square root to the variance to get the standard deviation
s = standard deviation
v = variance
s = sqrt(v)
s = sqrt(5.43)
s = 2.330326 which is approximate
s = 2.33
I rounded to 2 decimal places as this is the number of decimal places used for the variance.
Answer: the first part of this question is C, the second part of this question is D
