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2 years ago

If you roll a dice 6 times what is the probability that at least one 6

Mathematics

1 answer:

solong [7]2 years ago

6 0

The answer to the question is 1/6

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What is the difference between when a function is positive and when it is negative?

earnstyle [38]

When thinking about reinforcement, always remember that the end result is to try to increase the behavior, whereas punishment procedures are used to decrease behavior. For positive reinforcement, try to think of it as adding something positive in order to increase a response.

8 0

3 years ago

The portion of the parabola y²=4ax above the x-axis, where is form 0 to h is revolved about the x-axis. Show that the surface ar

castortr0y [4]

Answer:

See below for Part A.

Part B)

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

Step-by-step explanation:

Part A)

The parabola given by the equation:

$y^2=4ax$

From 0 to h is revolved about the x-axis.

We can take the principal square root of both sides to acquire our function:

$y=f(x)=\sqrt{4ax}$

Please refer to the attachment below for the sketch.

The area of a surface of revolution is given by:

$\displaystyle S=2\pi\int_{a}^{b}r(x)\sqrt{1+\big[f^\prime(x)]^2} \,dx$

Where r(x) is the distance between f and the axis of revolution.

From the sketch, we can see that the distance between f and the AoR is simply our equation y. Hence:

$r(x)=y(x)=\sqrt{4ax}$

Now, we will need to find f’(x). We know that:

$f(x)=\sqrt{4ax}$

Then by the chain rule, f’(x) is:

$\displaystyle f^\prime(x)=\frac{1}{2\sqrt{4ax}}\cdot4a=\frac{2a}{\sqrt{4ax}}$

For our limits of integration, we are going from 0 to h.

Hence, our integral becomes:

$\displaystyle S=2\pi\int_{0}^{h}(\sqrt{4ax})\sqrt{1+\Big(\frac{2a}{\sqrt{4ax}}\Big)^2}\, dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax}\Big(\sqrt{1+\frac{4a^2}{4ax}}\Big)\,dx$

Combine roots;

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax\Big(1+\frac{4a^2}{4ax}\Big)}\,dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax+4a^2}\, dx$

Integrate. We can consider using u-substitution. We will let:

$u=4ax+4a^2\text{ then } du=4a\, dx$

We also need to change our limits of integration. So:

$u=4a(0)+4a^2=4a^2\text{ and } \\ u=4a(h)+4a^2=4ah+4a^2$

Hence, our new integral is:

$\displaystyle S=2\pi\int_{4a^2}^{4ah+4a^2}\sqrt{u}\, \Big(\frac{1}{4a}\Big)du$

Simplify and integrate:

$\displaystyle S=\frac{\pi}{2a}\Big[\,\frac{2}{3}u^{\frac{3}{2}}\Big|^{4ah+4a^2}_{4a^2}\Big]$

Simplify:

$\displaystyle S=\frac{\pi}{3a}\Big[\, u^\frac{3}{2}\Big|^{4ah+4a^2}_{4a^2}\Big]$

FTC:

$\displaystyle S=\frac{\pi}{3a}\Big[(4ah+4a^2)^\frac{3}{2}-(4a^2)^\frac{3}{2}\Big]$

Simplify each term. For the first term, we have:

$\displaystyle (4ah+4a^2)^\frac{3}{2}$

We can factor out the 4a:

$\displaystyle =(4a)^\frac{3}{2}(h+a)^\frac{3}{2}$

Simplify:

$\displaystyle =8a^\frac{3}{2}(h+a)^\frac{3}{2}$

For the second term, we have:

$\displaystyle (4a^2)^\frac{3}{2}$

Simplify:

$\displaystyle =(2a)^3$

Hence:

$\displaystyle =8a^3$

Thus, our equation becomes:

$\displaystyle S=\frac{\pi}{3a}\Big[8a^\frac{3}{2}(h+a)^\frac{3}{2}-8a^3\Big]$

We can factor out an 8a^(3/2). Hence:

$\displaystyle S=\frac{\pi}{3a}(8a^\frac{3}{2})\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Simplify:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Hence, we have verified the surface area generated by the function.

Part B)

We have:

$y^2=36x$

We can rewrite this as:

$y^2=4(9)x$

Hence, a=9.

The surface area is 1000. So, S=1000.

Therefore, with our equation:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

We can write:

$\displaystyle 1000=\frac{8\pi}{3}\sqrt{9}\Big[(h+9)^\frac{3}{2}-9^\frac{3}{2}\Big]$

Solve for h. Simplify:

$\displaystyle 1000=8\pi\Big[(h+9)^\frac{3}{2}-27\Big]$

Divide both sides by 8π:

$\displaystyle \frac{125}{\pi}=(h+9)^\frac{3}{2}-27$

Isolate term:

$\displaystyle \frac{125}{\pi}+27=(h+9)^\frac{3}{2}$

Raise both sides to 2/3:

$\displaystyle \Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}=h+9$

Hence, the value of h is:

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

8 0

2 years ago

Read 2 more answers

when a pen is sold at a discount of 15%.There is a gain of rs 10.But if it is sold at 25% discount there is a loss of rs 2.Find

Svetradugi [14.3K]

When a pen is sold at a discount of 15% there is a gain of rs 10

When the pen is sold at 25% discount there is a loss of rs 2

Therefore the mark price of the pen can be calculated as follows

15% = c.p + 10 rs

25 % = c.p - 2 rs

25%-15% = 10 rs +2 rs

10% = 12 rs

10/100 = 12 rs

0.1= 12 rs

12/0.1

= 120 rs

Hence the mark price of the pen is 120 rs

6 0

3 years ago

Read 2 more answers

Which of the following is equal to 82 ? A.82.1116 B.82.6875 C.82.82 D.82.1611

wel

Step-by-step explanation:

4 0

3 years ago

How many ?!!!! ?!!! ?!!

mario62 [17]

hope this helps you alot☺✌❤

5 0

3 years ago