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Keith_Richards [23]
3 years ago
6

The house Trevor's family lives in has 6 people (including Trevor) and 3 bathrooms. In the past month, each person showered for

an average of 480 minutes and used an average of 72 litres of shower water (over the entire month). Water costs 0.20$ per litre. How much money did Trevor's family spend, in total, on shower water in the past month?
Mathematics
1 answer:
rjkz [21]3 years ago
6 0

Answer:

The amount spent on shower water by Trevor's family in the past month is $86.4

Step-by-step explanation:

The given information are;

The number of people in Trevor's family house are 6

The volume of water each person showered with during the month = 72 litres

The water costs = $0.20 per litre

The total volume of water used for shower by the 6 people = 72 × 6 = 432 litres

The amount spent on shower water by Trevor's family in the past month = The total volume of water used for shower × The cost per litre of water

The amount spent on shower water by Trevor's family in the past month = 432 × 0.2 = $86.4.

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The choice that combines the given sentences most effectively is: Option C : furnished: their.

<h3>What is sentence correction?</h3>

Sentence correction or sentence improvement is a type of grammatical practice where a sentence is given with a word or a phrase that requires grammatical changes or improvement.

Now, here in the given sentence:

"The plainer rooms are more sparsely <u>furnished. Their</u> architectural features, furnishings, and decorations are just as true to the periods they represent",

The most suitable choice, by sentence correction, to combine the two sentences will be: Furnished: their (option C).

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2 years ago
Sharon's Floral Shop needs to mail 959 checks to the bank. If they can put 3 checks in each envelope, how many checks will be in
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2 checks in the final envelope

Step-by-step explanation:

359 divided by three is 319,6

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I hope this was helpful!

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2 years ago
Read 2 more answers
Factor –7x3 21x2 3x – 9 by grouping. what is the resulting expression? (3 – 7x)(x2 – 3) (7x – 3)(3 x2) (3 – 7x2)(x – 3) (7x2 – 3
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The factors of the expression are \rm(3 - 7x^2)(x - 3)

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First group the first two terms and the last two terms together respectively.

The factors of the expression are given by;

\rm= -7x^3 + 21x^2 + 3x - 9\\\\=-7x^2(x - 3) + 3(x - 3)\\\\=(3 - 7x^2)(x - 3)

Hence, the factors of the expression are \rm \rm(3 - 7x^2)(x - 3).

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8 0
2 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
aliya0001 [1]

Answer:

A=1500-1450e^{-\dfrac{t}{250}}

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

R_{in} =(concentration of salt in inflow)(input rate of brine)

=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}

R_{out}=(concentration of salt in outflow)(output rate of brine)

=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}

Now, the rate of change of the amount of salt in the tank

\dfrac{dA}{dt}=R_{in}-R_{out}

\dfrac{dA}{dt}=6-\dfrac{A}{250}

We solve the resulting differential equation by separation of variables.  

\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}

Taking the integral of both sides

\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}

Recall that when t=0, A(t)=50 (our initial condition)

50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}

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Step-by-step explanation:

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