Please help with driving test!

Does anybody know how to do time with exponential decay

From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

$b_n=0.12\times b_{n-1}$

Why does this work? Initially, you start with 500 mL of air that you breathe in, so $b_1=500\text{ mL}$ . After the second breath, you have 12% of the original air left in your lungs, or $b_2=0.12\timesb_1=0.12\times500=60\text{ mL}$ . After the third breath, you have $b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}$ , and so on.

You can find the amount of original air left in your lungs after $n$ breaths by solving for $b_n$ explicitly. This isn't too hard:

$b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots$

and so on. The pattern is such that you arrive at

$b_n=0.12^{n-1}b_1$

and so the amount of air remaining after $50$ breaths is

$b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}$

which is a very small number close to zero.

5 0

3 years ago

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro

VladimirAG [237]

Answer:

Margin of error for a 95% of confidence intervals is 0.261

Step-by-step explanation:

Step1:-

Sample n = 81 business students over a one-week period.

Given the population standard deviation is 1.2 hours

Confidence level of significance = 0.95

Zₐ = 1.96

Margin of error (M.E) = $\frac{Z_{\alpha }S.D }{\sqrt{n} }$

Given n=81 , σ =1.2 and Zₐ = 1.96

Step2:-

 $Margin of error (M.E) = \frac{Z_{\alpha }S.D }{\sqrt{n} }$ 

 $Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }$ 

On calculating , we get

Margin of error = 0.261

Conclusion:-

Margin of error for a 95% of confidence intervals is 0.261

4 0

3 years ago

Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (−1, 0).

Hitman42 [59]

The two points that are most distant from (-1,0) are exactly (1/3, 4sqrt(2)/3) and (1/3, -4sqrt(2)/3) approximately (0.3333333, 1.885618) and (0.3333333, -1.885618) Rewriting to express Y as a function of X, we get 4x^2 + y^2 = 4 y^2 = 4 - 4x^2 y = +/- sqrt(4 - 4x^2) So that indicates that the range of values for X is -1 to 1. Also the range of values for Y is from -2 to 2. Additionally, the ellipse is centered upon the origin and is symmetrical to both the X and Y axis. So let's just look at the positive Y values and upon finding the maximum distance, simply reflect that point across the X axis. So y = sqrt(4-4x^2) distance is sqrt((x + 1)^2 + sqrt(4-4x^2)^2) =sqrt(x^2 + 2x + 1 + 4 - 4x^2) =sqrt(-3x^2 + 2x + 5) And to simplify things, the maximum distance will also have the maximum squared distance, so square the equation, giving -3x^2 + 2x + 5 Now the maximum will happen where the first derivative is equal to 0, so calculate the first derivative. d = -3x^2 + 2x + 5 d' = -6x + 2 And set d' to 0 and solve for x, so 0 = -6x + 2 -2 = -6x 1/3 = x So the furthest point will be where X = 1/3. Calculate those points using (1) above. y = +/- sqrt(4 - 4x^2) y = +/- sqrt(4 - 4(1/3)^2) y = +/- sqrt(4 - 4(1/9)) y = +/- sqrt(4 - 4/9) y = +/- sqrt(3 5/9) y = +/- sqrt(32)/sqrt(9) y = +/- 4sqrt(2)/3 y is approximately +/- 1.885618

7 0

3 years ago

What are the solutions to the equation y2 – 1 = 48?

Furkat [3]

${y}^{2} - 1 = 48 \\ {y}^{2} = 49 \\ y = \sqrt{49} \\ y = - 7 \: or \: 7$

${y}^{2} - 1 = 24 \\ {y}^{2} = 25 \\ y = \sqrt{25} \\ y = - 5 \: or \: 5$

Hope this helps. - M

6 0

3 years ago

What is the fourth term in the arithmetic sequence 13,10,7

Anettt [7]

The fourth term in the arithmetic sequence is 4. This is because you are subtracting 3 each time, and you already have your first three numbers in the pattern, so all you have to do is subtract another 3 in order to get 4.

5 0

3 years ago