What is A + B?<br> A = [ 3.42 1]<br> B = [5 6 89]

A tank with capactity 500 gal originally contains 200 gal water with 100 lbs of salt mixed into it. Water containing 1 lb of sal

prisoha [69]

Answer:

The amount of salt in the tank at any moment $t$ is

$f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

The concentration of salt in the tank when it is at the point of overflowing is $0.968$ .

The theoretical limiting concentration of an infinite tank is $1$ lb per gallon.

Step-by-step explanation:

Let $f(t)$ be the amount of salt in the tank at any time $t.$

Then, its time rate of change, $f'(t)$ , by (balance law).

Since three gallons of salt water runs in the tank per minute, containing $1$ lb of salt, the salt rate is

$3.1=3$

The amount of water in the tank at any time $t$ is.

$200+(3-2)t=200+t,$

Now, the outflow is $2$ gal of the solution in a minute. That is $\frac 2{200+t}$ of the total solution content in the tank, hence $\frac 2{200+t}$ of the salt salt content $f(t)$ , that is $\frac{2f(t)}{200+t}$ .

Initially, the tank contains $100$ lb of salt,

Therefore we obtain the initial condition $f(0)=100$

Thus, the model is

$f'(t)=3-\frac{2f(t)}{200+t}, f(0)=100$

$\Rightarrow f'(t)+\frac{2}{200+t}f(t)=3, f(0)=100$

$p(t)=\frac{2}{200+t} \;\;\text{and} \;\;q(t)=3$ Linear ODE.

so, an integrating factor is

$e^{\int p dt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2$

and the general solution is

$f(t)(200+t)^2=\int q(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac 1{(200+t)^2}\int 3(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac c{(200+t)^2}+200+t$

Now using the initial condition and find the value of $c$ .

$100=f(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}$

$\Rightarrow c=-4000000=-4\times 10^6$

$\Rightarrow f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

is the amount of salt in the tank at any moment $t.$

Initially, the tank contains $200$ gal of water and the capacity of the tank is $500$ gal. This means that there is enough place for

$500-200=300$ gal

of water in the tank at the beginning. As concluded previously, we have one new gal in the tank at every minute. hence the tank will be full in $30$ min.

Therefore, we need to calculate $f(300)$ to find the amount of salt any time prior to the moment when the solution begins to overflow.

$f(300)=-\frac{4\times 10^6}{(200+300)^2}+200+300=-16+500=484$

To find the concentration of salt at that moment, divide the amount of salt with the amount of water in the tank at that moment, which is $500$ L.

$\text{concentration at t}=300=\frac{484}{500}=0.968$

If the tank had an infinite capacity, then the concentration would be

$\lim\limits_{t \to \infty} \frac{f(t)}{200+t}= \lim\limits_{t \to \infty}\left(\frac{\frac{3\cdot 10^6}{(200+t)^2}+(200+t)}{200+t}\right)$

$= \lim\limits_{t \to \infty} \left(\frac{4\cdot 10^6}{(200+t)^3}+1\right)$

$=1$

Hence, the theoretical limiting concentration of an infinite tank is $1$ lb per gallon.

3 0

4 years ago

The number line represents values for x. Which inequality BEST describes the included values? A) x < −4 B) x > −4 C) x ≤ −

NeX [460]

Answer:

x ≥ −4

The line is shaded to the right of −4 where the values are greater. The closed circle indicates that −4 is include

8 0

3 years ago

Read 2 more answers

What is 168 hours converted to days?

koban [17]

Answer:

7 days

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

dmitriy555 [2]

Answer:

The interval is [910.053; 959.946]

p-value 0.00596

Decision: Reject null hypothesis.

Step-by-step explanation:

Hello!

You need to make a 95% Confidence Interval for the population mean of scholarship examination scores for the freshman.

It is known to be μ= 900 and the assistant dean wants to test if it changed.

The study variable is:

X: -scholarship examination score of one applicant.

population variance is known as σ²= (180)²

Assuming that the variable has a normal distribution the formula for the interval is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

935 ± 1.96* $\frac{180}{\sqrt{200} }$

The interval is [910.053; 959.946]

To test if the examination scores have changed the hypothesis is:

H₀: μ = 900

H₁: μ ≠ 900

α: 0.05

To use a Confidence Interval the following conditions should be met:

1) Both the test and the interval should be made for the same parameter.

2) The hypothesis has to be two_tailed

3) Confidence level 1 - α and significance level α should be complementary.

To make the decision you have to see if the value given to the population mean in the null hypothesis is contained or not by the interval.

If the value is contained by the interval, you do not reject the null hypothesis.

If the value is not contained by the interval, then the decision is to reject the null hypothesis.

Since 900 is not contained by the 95% Confidence interval [910.05; 959.95], the decision is to reject the null hypothesis. This means that the scholarship examination scores of freshman applications have changed.

To calculate the p-value you have to know the value of the statstic under the null hypothesis:

Z= $\frac{935-900}{\frac{180}{\sqrt{200} } }$ = 2.749 ≅ 2.75

p-value is

P(Z<2.75) + P(Z>2.75)= P(Z<2.75) + (1 - P(Z<2.75))= 0.00298+ (1 - 0.9702= 0.00596

I hope it helps!

4 0

3 years ago

What are all the angles in math

Serjik [45]

if u mean what i think u mean the angles are the following...

acute

obtuse

right

straight!

4 0

3 years ago

What is A + B? A = [ 3.42 1] B = [5 6 89]