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soldi70 [24.7K]
3 years ago
5

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

r minute and the well-stirred solution runs out at the same rate. How long will it be until there are 23 pounds of salt in the tank? Answer: the amount of time until 23 pounds of salt remain in the tank is minutes.
Mathematics
1 answer:
adoni [48]3 years ago
5 0

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

\frac{dC}{dt}=Ci*Qi-Co*Qo

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

\frac{dC}{dt}=-Co*Qo

Rearranging the equation, it becomes

\frac{dC}{C}=-Qo*dt

Integrating both sides

\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\

The final equation for the concentration of salt at any given time is

C=exp^{-3*t-0.693}

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}=  -\frac{-0.266}{3}=0.088

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Answer:

a) \bar X=144.3

b) The 95% confidence interval would be given by (138.846;149.754)      

c) n=34

d) n=298

e) n=1190

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean with this formula:

\bar X = \frac{\sum_{i=1}^n x_i}{n}

\bar X=144.3 represent the sample mean  

\mu population mean (variable of interest)

\sigma=8.8 represent the population standard deviation

n=10 represent the sample size  

a) Find a point estimate of the population mean

We can calculate the sample mean with this formula:

\bar X = \frac{\sum_{i=1}^n x_i}{n}

\bar X=144.3

b) Construct a 95% confidence interval for the true mean score for this population.

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that t_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

144.3-1.96\frac{8.8}{\sqrt{10}}=138.846    

144.3+1.96\frac{8.8}{\sqrt{10}}=149.754

So on this case the 95% confidence interval would be given by (138.846;149.754)    

Part c

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =+20 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (5) we got:

n=(\frac{1.960(8.8)}{3})^2 =33.05 \approx 34

So the answer for this case would be n=34 rounded up to the nearest integer

Part d

n=(\frac{1.960(8.8)}{1})^2 =297.493 \approx 298

So the answer for this case would be n=298 rounded up to the nearest integer

Part e

n=(\frac{1.960(8.8)}{0.5})^2 =1189.97 \approx 1190

So the answer for this case would be n=1190 rounded up to the nearest integer

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