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3 years ago

A sweatshirt is marked down from $28 to $21. What is the percentage of change?

Mathematics

2 answers:

bezimeni [28]3 years ago

8 0

Answer:

Step-by-step explanation:

$28 = 100%

$7 = 25%

$21 = 75%

so, therefore

% change = 100% - 75%

= 25%

<h2>MARK ME BRAINLIST</h2>

notka56 [123]3 years ago

7 0

Answer:

-25.00% decrease, quite a deal

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A college professor wants to survey a sample of students taking her large lecture course. There are about 150

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Step-by-step explanation:

B is the only answer that will ensure she gets graduate students in the sample without getting too few or too many.

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3 years ago

What is 11/20 as a decimail please help

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$\frac{11}{20} = 0.55$

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3 years ago

if 1 large bus and two small buses equal 100 people and 1 small bus and 2 large buses equal 158. how big is one large bus?

Fofino [41]

Answer:

One large bus can carry 72 people

Step-by-step explanation:

I kid you not I was messing around with my calculator and I found 72 for the large busses and 14 for the small ones. By doing 72+72+14 you get 158 and if you do 14+14+72 you get 100. Hope this helps.

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3 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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4 years ago

A rectangular park that measures 245 yards by 420 yards has a sidewalk around it.

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