(1) The average ( arithematic mean) cost per book for the 12 books on a certain table is k dollars. If a book that costs 18 dollars is removed from the table and replaced by a book that costs 42 dollars, then in terms of k, what will be the average cost per book, in dollars, for the books on the table???
a) k+2 b) k-2 c) 12+24/k d)12-24/k e)12k-6
I chose B, why OA A???
(2) 10000^100 is equivalent to which of the following??
I (100^2)(100^100)
II 100^200
III 10^400
A) None B)I only C)III only D)II and III only E)I,II,III
I chose B why the hell OA is D
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???
a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00
Several math probs, pls help to explain me, 600+
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(2) 10000^100 is equivalent to which of the following??
I (100^2)(100^100)
II 100^200
III 10^400
A) None B)I only C)III only D)II and III only E)I,II,III
I chose B why the hell OA is D
10000^100 = (10^4)^100 = 10^400
I (100^2)(100^100) = (10^4)(10^200) = 10^204
II 100^200 = (10^2)^200 = 10^400
III 10^400
II and III are true
The correct answer is D.
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Cost/book - k
No of books - 12
Total cost - 12k
Now a $18 book is removed - (12k-18)
And, $42 book is added - (12k-18+42) = 12k+24
Avg = total/ No of books
= 12k+24/2 = k+2
No of books - 12
Total cost - 12k
Now a $18 book is removed - (12k-18)
And, $42 book is added - (12k-18+42) = 12k+24
Avg = total/ No of books
= 12k+24/2 = k+2
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In the first problem, you made an error in the sign (-18+42) = 24. Therefore average cost is (12k+24)/12 = k+2
In second problem ii 100^200 = (10^2)^200 = 10^400 . (10^a)^b = (10)^ab
For problem 3
The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM
In second problem ii 100^200 = (10^2)^200 = 10^400 . (10^a)^b = (10)^ab
For problem 3
The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM
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Wont the tank reach max when t=1?sl750 wrote: For problem 3
The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM
You then get; -20(1-5) + 500 = 80+500 = 580.
From answer choices earliest time would be max level, wont it be A?
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Yes the answer is A as it's the only time in the choice when (t-5) is lowest and -20(t-5) is lowest value in -ve.shankar.ashwin wrote:Wont the tank reach max when t=1?sl750 wrote: For problem 3
The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM
You then get; -20(1-5) + 500 = 80+500 = 580.
From answer choices earliest time would be max level, wont it be A?
So the value at 5:30 will be N(t)= -20(5.5-5)+500 = -10+500 = 490
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For no. 3
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???
a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00
Could someone explain in details , tahnk you.
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???
a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00
Could someone explain in details , tahnk you.
- sl750
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Ok, you make a point. But how does that make t=5:30?, if t=1, then the earliest time would be 3:00 AM and not 5:30 AMshankar.ashwin wrote:Wont the tank reach max when t=1?sl750 wrote: For problem 3
The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM
You then get; -20(1-5) + 500 = 80+500 = 580.
From answer choices earliest time would be max level, wont it be A?
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Can you confirm if the equation is correct? I suspect it should read N(t) = -20(t-5)^2 + 500tracyyahoo wrote:For no. 3
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???
a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00
Could someone explain in details , tahnk you.
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No, I just gave an example as t=1. for 5.30 am, it woud obviously be 3.5 and still the depth would be higher than at 7am according to the equation.
sl750 wrote:Can you confirm if the equation is correct? I suspect it should read N(t) = -20(t-5)^2 + 500tracyyahoo wrote:For no. 3
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???
a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00
Could someone explain in details , tahnk you.