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3 years ago

8

The parking lot at a store has a width of 20 yards 2 feet and the length of 30 yards. The cost to repave the parking lot is two

dollars per square foot. Explain how much it would cost to repave the parking lot.

2 answers:

IceJOKER [234]3 years ago

7 0

The cost to repave the parking lot is $912

Ronch [10]3 years ago

3 0

The cost would end up being $912

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3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

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3 years ago

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Answer:

5.41127

Step-by-step explanation:

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3 years ago

What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean

Elena-2011 [213]

Answer:

0.658 is the probability that a sample 90 test takers will provide a sample mean test score within 10 points of the population mean of 502.

Step-by-step explanation:

The following information is missing:

The standard deviation of population is 100.

We are given the following information in the question:

Population mean, μ = 502

Standard Deviation, σ = 100

Sample size, n = 90

Standard error =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{100}{\sqrt{90}} = 10.54$

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(test score within 10 points)

$P(492 \leq x \leq 512) \\\\= P(\displaystyle\frac{492 - 502}{10.54} \leq z \leq \displaystyle\frac{512-502}{10.54}) \\\\= P(-0.9487 \leq z \leq 0.9487)\\= P(z \leq 0.9487) - P(z < -0.9487)\\= 0.829 -0.171 = 0.658 = 65.8\%$

$P(492 \leq x \leq 512) = 65.8\%$

0.658 is the probability that a sample 90 test takers will provide a sample mean test score within 10 points of the population mean of 502.

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3 years ago