Find the length of the side of a square whose area is 57.1536

A random walker is observed to take a total of N steps, n of which are to the right (and the rest n^' are to the left). Suppose

ioda

Answer:

p = 0.635

Step-by-step explanation:

A step to the right is n and on ten different days, it was recorded to be

14, 13, 11, 12, 11, 12, 16, 16, 14, 8.

n = average of all the steps,

The average is the sum of variables divided by the number of variables

Average = (14+13+11+12+11+12+16+16+14+8)/10

Average = 12.7 steps.

σₙ² = variance of n

Variance is an average of the squared deviations from the mean.

[Σ(x - xbar)²/N]

Σ(x - xbar)² = (14-12.7)² + (13- 12.7² + (11-12.7)² + (12-12.7)² + (11-12.7)² + (12-12.7)² + (16-12.7)² + (16-12.7)² + (14-12.7)² + (8-12.7)² = 54.1

Variance = σₙ² = (54.1/10) = 5.41

standard deviation = σₙ = √(variance) = √5.41 = 2.326.

The proportion, p can be computed in two ways.

Mean = Np or standard deviation = √[Np(1-p)]

where N = total number of steps = 20

Mean = 12.7 = 20p

p = (12.7/20) = 0.635

Or

Standard deviation = 2.326 = √[20p(1-p)]

5.41 = 20p - 20p²

20p² - 20p + 5.41 = 0

Solving the quadratic equation,

p = 0.5 + 0.143i or p = 0.5 - 0.143i

not a unique or feasible solution, hence, the probability of a step to the right remains p = 0.635

Hope this Helps!!!

5 0

3 years ago

What is 3 numbers between 2.15 and 2.17

hoa [83]

2.153, 2.164, 2.167
Those are just a 3 of thousands of possibilities.

7 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

Which of the following is a point slope equation of the line below (10,5)(2,2)

salantis [7]

Point slope form follows the equation y-y₁=m(x-x₁), so we want it to look like that. Starting off with m, or the slope, we can find this using your two points with the formula $\frac{ y_{1} -y_{2} }{ x_{1} - x_{2} }$ . Note that y₁ and x₁ are from the same point, but it does not matter which point you designate to be point 1 and point 2. Thus, we can plug our numbers in - the x value comes first in the equation, and the y value comes second, so we have
$\frac{5-2}{10-2} = \frac{3}{5}$ as our slope. Keeping in mind that it does not matter which point is point 1 and which point is point 2, we go back to y-y₁=m(x-x₁) and plug a point in (I'll be using (10,5)). Note that x₁, m, and y₁ need to be plugged in, but x and y stay that way so that you can plug x or y values into the formula to find where exactly it is on the line. Thus, we have our point slope equation to be $y- 5= \frac{3}{5} (x-10)$

Feel free to ask further questions!

7 0

3 years ago

Select the correct answer.

Sonbull [250]

Answer:

B.

The graph is stretched vertically and shifted to the left 1 unit.

4 0

2 years ago