Answer:
a
Step-by-step explanation:
i just took the test
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.
Answer:
68,000,000
Step-by-step explanation:
Look at the hundred thousands place. If it's a 5 or higher you round up and if it's lower you round down. Since there is 4 in the hundred thousands place you round down. Everything to the right turns into a zero.
95% of red lights last between 2.5 and 3.5 minutes.
<u>Step-by-step explanation:</u>
In this case,
- The mean M is 3 and
- The standard deviation SD is given as 0.25
Assume the bell shaped graph of normal distribution,
The center of the graph is mean which is 3 minutes.
We move one space to the right side of mean ⇒ M + SD
⇒ 3+0.25 = 3.25 minutes.
Again we move one more space to the right of mean ⇒ M + 2SD
⇒ 3 + (0.25×2) = 3.5 minutes.
Similarly,
Move one space to the left side of mean ⇒ M - SD
⇒ 3-0.25 = 2.75 minutes.
Again we move one more space to the left of mean ⇒ M - 2SD
⇒ 3 - (0.25×2) =2.5 minutes.
The questions asks to approximately what percent of red lights last between 2.5 and 3.5 minutes.
Notice 2.5 and 3.5 fall within 2 standard deviations, and that 95% of the data is within 2 standard deviations. (Refer to bell-shaped graph)
Therefore, the percent of red lights that last between 2.5 and 3.5 minutes is 95%
Answer:
24/25
Step-by-step explanation:
0.96%
96/100
24/25