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aniked [119]
2 years ago
11

Water fills a tank at a rate of 150 litres during the first hour, 350 litres during the second, 550 litres during the third and

so on. Find the number of hours necessary to fill a rectangular tank 16m x 7m x 7m.

Mathematics
2 answers:
Lubov Fominskaja [6]2 years ago
5 0
Volume of tank = (16m)(7m)(7m) = 784 m³

Conversion of m³ to L:
(784 m³) × (1000L / 1m³) = 784,000 L

Rate in the 1st hour:
150 liters/hr

Rate in the 2nd hour:
350 liters/hr

Rate in the 3rd hour:
550 liters/hr

It is apparent that the Fill Rate is increasing by 200 liters/hr every subsequent hour . . . so that can be represented by the following equation

where:
t = number of hours

Fill rate (i.e. volume of water filled into tank within the specified hour) = 150 + 200(t - 1)

For t = 1 . . . Fill rate = 150 L/hr
For t = 2 . . . Fill rate = 350 L/hr
For t = 3 . . . Fill rate = 550 L/hr

Because after every hour there has been more water added to the tank, this problem can be represented as a geometric sequence in order to account for the compounding of the volume after each time step, but it can also be tabulated (which seems to me to be the more direct/simple approach), so I will build a table that accounts for the increasing Fill Rate and the compounding of water volume after each time step . . .

(see attached)

The answer (after all of this) is . . .  t = 88 hrs 17 1/2 mins (approx)



Alborosie2 years ago
3 0
Putting this as an arithmetic sequence gives:

u_n = 150+200(n-1)

The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L

The sum of an arithmetic series can be written as:

S_n=n/2 [2a+(n-1)d] = 784 000
\\n/2[2(150)+(n-1)200] = 784 000
\\n[300+200(n-1)=1 568 000
\\300n+200n^2-200n = 1 568 000
\\200n^2+100n- 1 568 000 = 0
\\2n^2 +n- 15680 = 0

\\n= 88.2...,-88.7

n has to be positive, so we get

n = <u>88.2 hours (3 s.f.)</u>
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Virty [35]

Answer:

The year 1996

With population of both 21600

Step-by-step explanation:

From 1990 to 2000 = 10 years

So city A grew from 12000 to 28000 that is city A had an increase of 16000 in 10 years.

While city b grew from 18000 to 24000 , that's an increase of 6000 in 10 years to.

For city A

10 years= 16000

1 year = 16000/10

1 year = 1600

For city B

10 years = 6000

1 year = 6000/10

1 year = 600

So we are to find what year the both cities had same population.

12000 + x1600 = y

18000 + x600 = y

X is the year difference

Y is the population at that year

Eliminating y gives

6000= x1000

X= 6

If x is 6

18000+3600= y

21600= y

So 6 years + 1990 = 1996

4 0
2 years ago
A book shelf holds twenty books. Eleven are hard- covered, nine of the books are nonfiction, and four of the nonfiction books ar
ololo11 [35]

Answer:

1) Find the probability that a chosen book is nonfiction and hard-covered. P(N ∩ H) = 4/20

2) Find the probability that a chosen book is hard-covered. P(H) =  11/20

3) Calculate the conditional probability. = 4/11

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1.) 4/20

2.) 11/20

3.) 4/11

7 0
2 years ago
Read 2 more answers
Four brothers each bought two hotdogs and a bag of chips at the concession stand.if the bag of chips was $1.25,and the total was
ahrayia [7]

Answer:

  $1.65

Step-by-step explanation:

The total purchase can be described by ...

  4(2h +1.25) = 18.20 . . . . where h is the price of a hot dog

  8h = 13.20 . . . . . . . . . . . subtract 5.00

  h = 1.65 . . . . . divide by 8

The price of a hot dog was $1.65.

5 0
2 years ago
Jason hires groundskeepers. They work an average of 60 hours per month. If seven groundskeepers are hired at $12/hour, what is t
Kamila [148]
Multiply the cost per hour by 60 to find the cost for a groundskeeper
cost = 12 × 60
cost = 720
The cost for each groundskeeper is $720

Find the cost for seven groundkeepers
Multiply cost for each groundskeeper by 7
cost = 720 × 7
cost = 5,040

The monthly labor cost for seven groundskeepers is $5,040
5 0
2 years ago
In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. A long-distance telephone c
scoundrel [369]

Answer:

We know that In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. And we want to test if the mean duration of long-distance phone calls has changed from the 1990 mean of 7.2 minutes (alternative hypothesis) and the complement rule would represent the null hypothesis.

The correct system of hypothesis are:

Null hypothesis: \mu =7.2

Alternative hypothesis: \mu \neq 7.2

So then the best option for this case would be:

H0: μ = 7.2 minutes Ha: μ ≠ 7.2 minutes

Step-by-step explanation:

We know that In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. And we want to test if the mean duration of long-distance phone calls has changed from the 1990 mean of 7.2 minutes (alternative hypothesis) and the complement rule would represent the null hypothesis.

The correct system of hypothesis are:

Null hypothesis: \mu =7.2

Alternative hypothesis: \mu \neq 7.2

So then the best option for this case would be:

H0: μ = 7.2 minutes Ha: μ ≠ 7.2 minutes

And in order to test the hypothesis we can use a one sample t test or z test depending if we know the population deviation or not

4 0
2 years ago
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