What is the probability of getting an even number of heads when flipping an unfair coin n times?

Please help me. ASAP.Please I need answer step-by-step.

nikitadnepr [17]

so this is how i was taught the inverse function. f(x) is basicly the same thing as saying y

so y=x over x+3

you switch the top x with y so it turns in to x=y over x+3

the problem should tell you what x equals so substitute x with the number it equals and then solve

here is an example f(x)=x over x+3 and x=14

x=y over x+3

14=y over 14+3

the inverse inverse function would probably be just solve the equation so just substitute the number they give you for x in for x

4 0

3 years ago

a baker is making bread dough.He uses 3 cups of flour for every 8 ounces of water. How many cups of flour will be use if he used

Nataliya [291]

Answer:

36 cups of flour

Step-by-step explanation:

if 3 cups of flour=8 ounces.what about 96 onces you cross multiply

7 0

3 years ago

Can y’all help me a question 25?!

Elenna [48]

Answer:

19.8839 miles away

Step-by-step explanation:

:)

3 0

2 years ago

MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) t

ioda

Answer:

We would have

$l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}$

where " l " is length, " w" is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

Surface area for a box with no top = $lw+2lh+2wh = 64$

where " l " is length, " w" is width and "h" is height.

Step 2.

Remember as well that

Volume of the box = $l*w*h$

Step 3

We can now use lagrange multipliers. Lets say,

$F(l,w,h) = lwh$

and

$g(l,w,h) = lw+2lh+2wh = 64$

By the lagrange multipliers method we know that

$\nabla F = \lambda \nabla g$

Step 4

Remember that

$\nabla F = (wh,lh,lw)$

and

$\nabla g = (w+2h,l+2h , 2w+2l)$

So basically you will have the system of equations

$wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)$

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

$lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)$

Then you would get

$l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)$

You can get rid of $\lambda$ from these equations and you would get

$lw+2lh = lw+2wh = 2wh+2lh$

And from those equations you would get

$l = w =2h$

Now remember the original equation

$lw+2lh+2wh = 64$

If we plug in what we just got, we would have

$l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}$

7 0

3 years ago

A circle is centered at (2, -5) and has radius 3. Write its equation in standard form. A) (x + 2)2 + (y - 5)2 = 3 B) (x - 2)2 +

ki77a [65]

D.
$(x - h)^{2} + {(y - k)}^{2} = r^{2}$
Where (h, k) is the center of the circle and the radius is squared.

4 0

3 years ago

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