The current of each math work and the percentage of

(a) If you toss 10 coins, what percent of the time will you get the three most likely macrostates (6 heads and 4 tails, 5 heads

FromTheMoon [43]

This question is incomplete, the microstates table is left out which will be uploaded along this answer;

Answer:

a) the required is 65.6%

b) the required time is 4.2666 hrs

Step-by-step explanation:

using the microstates from the table in the diagram;

a)

the percent times the above any of the above microstates occurs is;

% (6 heads and 4 tails, 5 heads and 5 tails, 4 heads and 6 tails)

= ((210 + 252 + 210)/1024) × 100

= 0.6562 × 100

= 65.6%

Therefore the required is 65.6%

b)

the probability of getting 10 heads and 0 tail or 0 tail head and 10 tails is;

p( 10H, 0T or 0H, 10T = 2/1024

= 1/512

so on average, it will take 512 trails or at a rate of 2 trials/min, 256 mim in hours;

(256 min) ( 1h/60min) = 4.2666 hrs

Therefore the required time is 4.2666 hrs

8 0

2 years ago

What is y(x)+ab-5 I don’t get the problem

diamong [38]

Answer:

ab+xy−5

Step-by-step explanation:

yx+ab−5 = xy+ab+−5

Simplify-

ab+xy−5

6 0

3 years ago

Solve 13x - 4 < 12(x - 1).

Effectus [21]

13x - 4 < 12(x-1)
13x - 4 < 12x - 12
13x - 12x < -12 + 4
x < -8
x > 8

x is greater than 8

6 0

3 years ago

Find the distance covered in 10 revolution by a road roller of diameter 4.2m

IceJOKER [234]

The distance is 131.9m.

3 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago