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2 years ago

5

suppose we toss 100 fair coins and record the percentage of heads. according to the central limit theorem, the percentages of he

ads resulting from such experiments are approximately normally distributed. Find the mean and the standard deviation of this normal distribution.

1 answer:

SVETLANKA909090 [29]2 years ago

6 0

answer is attached below

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Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth

taurus [48]

Answer:

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

$y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\$

It is a homogeneous solution:

$y_h(t)=c_1e^{-2i t}+c_2e^{2i t}$

Now, we finding a particular solution.

$y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\$

we get

$y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

7 0

3 years ago

How do you this question? Please explain.

Nuetrik [128]

Answer:

(C) 2√15

Step-by-step explanation:

Recognize that all the triangles are right triangles, so are similar to each other. In these similar triangles, the ratio of the short side to the long side is the same for all.

... CB/CA = CT/CB

... CB² = CA·CT = 10·6 = 60 . . . . . . . . . . multiply by CA·CB; substitute values

... CB = √60 = 2√15 . . . . . . . take the square root; simplify

_____

<em>Comment on this solution</em>

The altitude to the hypotenuse of a right triangle (CB in this case) divides the hypotenuse into lengths such that the altitude is their geometric mean. That is ...

... CB = √(AC·CT) . . . . as above

This is true for any right triangle — another fact of geometry to put in your list of geometry facts.

3 0

2 years ago

For what value of c does x^2−2x−c=4 have exactly one real solution?<br> PLEASE HELP!!!!!!!

Paul [167]

Answer:

-5

Step-by-step explanation:

Moving all terms of the quadratic to one side, we have

$x^2-2x-(c+4)=0$ .

A quadratic has one real solution when the discriminant is equal to 0. In a quadratic $ax^2+bx+d$ , the discriminant is $\sqrt{b^2-4ad}$ .

(The discriminant is more commonly known as $\sqrt{b^2-4ac}$ , but I changed the variable since we already have a $c$ in the quadratic given.)

In the quadratic above, we have $a=1$ , $b=-2$ , and $d=-(c+4)$ . Plugging this into the formula for the discriminant, we have

$\sqrt{(-2)^2-4(1)(-(c+4))$ .

Using the distributive property to expand and simplifying, the expression becomes

$\sqrt{4-4(-c-4)}=\sqrt{4+4c+16}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{20+4c}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{4}\cdot\sqrt{5+c}\\~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{c+5}.$

Setting the discriminant equal to 0 gives

$2\sqrt{c+5}=0$ .

We can then solve the equation as usual: first, divide by 2 on both sides:

$\sqrt{c+5}=0$ .

Squaring both sides gives

$c+5=0$ ,

and subtracting 5 from both sides, we have

$\boxed{c=-5}.$

3 0

2 years ago

rock is thrown upward with a velocity of 27 meters per second from the top of a 23 meter high cliff and it misses the cliff on t

ahrayia [7]

the rock will be at 11 meters from the ground level after 5.92 seconds

Step-by-step explanation:

The motion of the rock is a free-fall motion, since the rock is acted upon the force of gravity only. Therefore, it is a uniformly accelerated motion, so its position at time t is given by the equation:

$y=h+ut+\frac{1}{2}at^2$

where

h = 23 m is the initial height

u = 27 m/s is the initial velocity, upward

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

t is the time

We want to find the time t at which the position of the rock is

y = 11 m

Substituting and re-arranging the equation, we find

$11=23+27t-4.9t^2\\4.9t^2-27t-12=0$

This is a second-order equation, which has solutions:

$t=\frac{27\pm \sqrt{(-27)^2-4(4.9)(-12)}}{2(4.9)}=\frac{27\pm \sqrt{964.2}}{9.8}$

So

$t_1 = -0.41 s$

$t_2=5.92 s$

The first solution is negative so we neglect it: therefore, the rock will be at 11 meters from the ground level after 5.92 seconds.

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

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8 0

3 years ago

(05.03)An irregular polygon is shown below:

SVEN [57.7K]

9units sorry for not showing work though but glad I can help

7 0

3 years ago