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3 years ago

What in math what does it mean when I'd a sample process is in control or not?

1 answer:

8 0

I'm not totally sure what you are asking, but maybe this is a question about dependent and independent variables?

If so, these types of variables are used in math and science as measures. In math, independent variables and dependent variables are the "tools" used in an experiment.

Independent variables are the variables that are controlled during the experiment. These variables do not change during the experiment.

Dependent variables are the variables whose values change based on the independent variables. These variables are the "outcome" portion.

I have no idea if this answers your question or not. All I can say is that the "controlled" part of a sample would be the independent variable.

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The population of Adamsville grew from 7,000 to 12,000 in 5 years. Assuming uninhibited exponential growth, what is the expected

mr Goodwill [35]

We have that if we assume standard exponential growth, the equation of the population will be: $7000*e^{kt}$ if we start counting from the moment that the population was 7000. We are given that 7000* $e^{5k}$ =12000, namely that P(5)=12000 and we need to find $7000e^{(5+5)k}=7000e^{10k}$ . Since e^(10k)=e^(5k)*e^(5k), and since we can solve for e^(5k) from P(5), we have:
e^(5k)=12000/7000 and we can calculate P(10). P(10)=7000 $\frac{12000^2}{7000^2}$ = 20571 people.

3 0

3 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago

Please help me with this.

kompoz [17]

Answer:

$75\sqrt{2}(\sqrt{3}-1)$

Step-by-step explanation:

In every right triangle, a leg is given by the multiplication of the sine of its opposite angle and the hypothenuse.

In this case, the opposite angle of x is 15°, and the hypothenuse is 300. So, we have

$x = 300\sin(15) = 75\sqrt{2}(\sqrt{3}-1)$

3 0

3 years ago

What is 4/5×24/12 equal to and simplified

klasskru [66]

It should be 8/5 cause 4/5 times 24/12= 4/5 times 2/1. then u multiply.

4 0

3 years ago

A farmer had twice as many chickens as ducks on his farm. After he sold 166 chickens and ducks, he had half as many chickens as

mamaluj [8]

Answer:

The farmer have at first 202 chickens.

Step-by-step explanation:

Given:

A farmer had twice as many chickens as ducks on his farm. After he sold 166 chickens and ducks, he had half as many chickens as ducks left.

Now, to find the chickens farmer have at first.

Let the chickens be $x.$

And, the ducks be $y.$

As, the farmer had twice as many chickens as ducks on his farm.

So, $x=2y$ ......(1)

As, given the farmer after selling 166 chickens and ducks, he had half as many chickens as ducks left.

According to question:

$x-166=(y-29)\times \frac{1}{2}$

$x-166=\frac{y-29}{2}$

Substituting the value of $x$ from equation (1) we get:

$2y-166=\frac{y-29}{2}$

By cross multiplying we get:

$4y-332=y-29$

Adding both sides 332 we get:

$4y=y+303$

Subtracting both sides by $y$ we get:

$3y=303$

Dividing both sides by 3 we get:

$y=101.$

Now, to get the number of chickens substituting the value of $y$ in equation (1):

$x=2y\\\\x=2\times 101\\\\x=202.$

Therefore, the farmer have at first 202 chickens.

7 0

3 years ago