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Troyanec [42]
3 years ago
12

A tank originally contains 160 gal of fresh water. Then water containing 1 4 lb of salt per gallon is poured into the tank at a

rate of 4 gal/min, and the mixture is allowed to leave at the same rate. After 8 min the process is stopped, and fresh water is poured into the tank at a rate of 6 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 8 min.
Mathematics
1 answer:
Anika [276]3 years ago
7 0

Answer:

245.86 lb of salt

Step-by-step explanation:

Let y(t) be the amount of salt in the tank at instant t. Then y'(t) represents the rate of change of the amount of salt in the tank at instant t.

The water entering the tank contains 14 lb of salt per gallon and enters at a speed of 4 gal/min, so every minute 14*4=56 lb of salt are entering the tank the first 8 min.

y'(t)=56\;(0\leq t\leq 8)

After 8 minutes, as the volume of tank keeps 160 gal., the concentration of salt is \frac{y(t)}{160}.

Since the water is now leaving the tank at 6 gal/min, the salt is leaving the tank at a speed of -6\frac{y(t)}{160}=-0.0375y(t).

Notice that the speed is negative because the salt is leaving the tank, so the amount of salt is decreasing.

Now we have the following ordinary differential equation

y'(t)=56 \; 0 \leq t \leq 8

-0.0375y(t) \; 8 \leq t \leq 16

Solving this differential equation we get a piecewise function defined as follow

y(t) = 56t + C \; (0\leq t\leq 8)

where C is a constant.

Since the tank had no salt at the beginning of the process. Y(0)=0, so

y(t) = 56t \; (0\leq t\leq 8)

For 0\leq t\leq 8 we have

y'(t)=-0.0375y(t)

and solving this equation we get

y(t)=Ce^{-0.0375t}

where C is a constant that can be found with the initial condition y(8) = 56*8 = 448

and our solution is

y(t)=56t \; 0 \leq t \leq 8

448e^{-0.0375t} \; 8\leq t\leq 16

To know what the amount of salt is at the end of the process, we must evaluate y(16), which is

y(16)=448e^{-0.0375*16}=448*0.5488\approx 245.867

So, at the end of the process the tank contains 245.867 lb of salt.

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Hey!

Finding the factorial for the first few numbers, we have:

1!=1

2!=2

3!=6

4!=24

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7!=720*7

8!=720*56

What we can see as a clear pattern from 5! and on is that our number ends with a 0, making the units digit 0. Therefore, when we add the units digit of 5! and on, we have a result of 0. So, we can simply add the units digits of 1!, 2!, 3!, and 4!, which is 1+2+6+4=13. Since the units digit is the last number, we can drop the tens place to get an answer of 3.


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X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
2 years ago
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