Answer:
It will take 6 hours for the new pump to drain the pool.
Step-by-step explanation:
As the complete question is not given, the complete question is found online and is attached herewith
Let the rate of new pump is given as x=W/t_1
Let the rate of the old pump is given as y=W/t_2
it is given that the time t_2=2t_1
So by substituting the values of t_2 in the rate equation of y
y=W/2t_1
y=(W/t_1*2)=x/2
Also the total rate of both the pumps is given as W/t3 where t3 is given as 4 hours so the equation becomes
x+y=W/4
x+x/2=W/4
3x/2=W/4
As x=W/t_1
3W/2t_1=W/4
Now as W is same on both sides so
3/2t_1=1/4
12=2t_1
t_1=6 hours
So it will take 6 hours for the new pump to drain the pool.
Answer:
The ratio of juice boxes remaining to juice boxes drunk = 20 / 60 = 1/3
Step-by-step explanation:
<h3>
Answer:</h3>
- B. f(x) = 3,000(0.85)^x
- $1566.02
<h3>
Step-by-step explanation:</h3>
Part A
At the end of the year, the value of the computer system is ...
... (beginning value) - 15% · (beginning value) = (beginning value) · (1 - 0.15)
... = 0.85 · (beginning value)
Since the same is true for the next year and the next, the multiplier after x years will be 0.85^x. Then the value after x years is ...
... f(x) = (beginning value) · 0.85^x
The beginning value is given as $3000, so this is ...
... f(x) = 3000·0.85^x
____
Part B
For x=4, this is ...
... f(4) = 3000·0.85^4 = 3000·0.52200625 ≈ 1566.02
The value after 4 years is $1566.02.