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2 years ago

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g A coin where the probability of heads is 0.3 is flipped 2000 times. Use the normal approximation to the binomial distribution

to find the probability of getting between 575 and 618 heads (inclusive).

1 answer:

matrenka [14]2 years ago

5 0

Title:

<h2>See the explanation.</h2>

Step-by-step explanation:

The coin is to be flipped 2000 times.

The probability of getting a head is 0.3.

It is given that, we need to get a head in between 575 and 618 times.

If we will get 575 heads then we will get (2000 - 575) tails.

Hence, if we take that we want to get n times head, then we will get (2000 - n) times tail.

The probability of not getting a head is (1 - 0.3) = 0.7.

The probability of getting n times head is $^{2000}C_n (0.3)^n \times (0.7)^{2000 - n}$ .

The required probability is ∑ $^{2000}C_n (0.3)^n \times (0.7)^{2000 - n}$ , where $575 \leq n \leq 618$ .

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A trough is 8 feet long and has perpendicular cross section in the shape of an isosceles triangle (point down) with base 1 foot

murzikaleks [220]

Answer:

Dh/dt = 0.082 ft/min

Step-by-step explanation:

As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of 1 feet and height h = 2 feet.

The volume of a circular cone is:

V(c) = 1/3 * π*r²*h

Then differentiating on both sides of the equation we get:

DV(c)/dt = 1/3* π*r² * Dh/dt (1)

We know that DV(c) / dt is 1 ft³ / 5 min or 1/5 ft³/min

and we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment

By proportion we know

r/h ( at the top of the cone 0,5/ 2) is equal to r/0.5 when water is 1/2 foot deep

Then r/h = 0,5/2 = r/0.5

r = (0,5)*( 0.5) / 2 ⇒ r = 0,125 ft

Then in equation (1) we got

(1/5) / 1/3* π*r² = Dh/dt

Dh/dt = 1/ 5*0.01635

Dh/dt = 0.082 ft/min

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3 years ago

Anuta_ua [19.1K]

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3 years ago

Read 2 more answers

Please help I will reward brainly

aalyn [17]

P(heads, and even and blue) = 1/2 *3/6 *1/4=1/2*1/2*1/4=1/16

P(heads, and 5, and green) = 1/2*1/6*1/4=1/48

P((heads or tail), and odd, and red)= (1/2+1/2)*3/6*1/4=1*1/2*1/4 = 1/8

P((heads or tail), and 3, and yellow)= (1/2+1/2)*1/6*1/4=1*1/6*1/4=1/24

P(tails, and prime, and not green) = 1/2 *3/6*3/4= 1/2*1/2*3/4=3/16
(Prime numbers here are 2,3 and 5)

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3 years ago

Which description is correct for the polynomial 2x^2 + 2?

julia-pushkina [17]

D. Quadratic Binomial

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3 years ago

Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

vodomira [7]

Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two <u>coincident</u><u> points</u>

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using <u>substitution</u><u> method</u>

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by <em>Pythagorean</em><em> theorem</em> we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by <em>Zero</em><em> product</em><em> </em><em>property</em> we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

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