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2 years ago

A con is tossed n times. How many difference sequences of head and tails can you get?

Mathematics

2 answers:

ryzh [129]2 years ago

5 0

Answer:

2^n

Step-by-step explanation:

So whenever you flip a coin, you can see it as 2 nodes branching off of each existing node. so for example when you flip a coin once you're going to have 2 sequences initially H and T, now when you flip a coin again for each of those 2 sequences 2 are going to branch off of that, making the total sequences 4, and the next flip 2 sequences are going to branch off each of the 4 sequences and so on. this can generally be described as: 2^n

I attached an image describing this a bit better but the bottom line is that for each 'end node'/sequence you're going to have 2 branch off of it, thus for each coin flip the number of sequences multiplies by 2

nataly862011 [7]2 years ago

4 0

Answer:

2^n

Step-by-step explanation:

dont worry bout it ur welcome

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WHERE WOULD THESE GOOO?

maksim [4K]

Answer:

√20 = 4.5

√30 = 5.5

√23 = 4.8

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3 years ago

Prove Sin3theta = 3sintheta - 4sin^3theta

trapecia [35]

Express the left hand side as

sin3theta=sin(theta+2theta)

now expand the right side of this equation using color(blue)"Addition formula"

color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))

rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))

The right hand side is expressed only in terms of sintheta's

so we use cos2theta=1-2sin^2theta........(1)

color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)

Replace cos2theta" and " sin2theta by the expansions (1) and (2)
into (A)

sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)

and expanding brackets gives.

sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))

Replace cos^2theta=1-sin^2theta" into (B)"

rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)

and expanding 2nd bracket gives.

sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta

Finally, collecting like terms.

sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"

3 0

3 years ago

How will the perimeter of the rectangle change if each side is increased by a factor of 10? The perimeter will be 1/10 the origi

goldfiish [28.3K]

Answer:

The perimeter will be 10 times the original.

Step-by-step explanation:

Formula for Perimeter of rectangle is

2l + 2 w. From the diagram shown, the perimeter is 2×6 +2×3 = 12+6 =18cm

If if each side is increased by a factor of 10

Our length becomes 6×10= 60cm

Our width becomes 3×10= 30cm

New perimeter becomes

2×60 + 2×30

= 120+60 = 180cm

New perimeter(180 cm)= 10× original perimeter (18cm)

So The perimeter will be 10 times the original.

4 0

3 years ago

Read 2 more answers

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Lerok [7]

Just divided 1.3/100 and it gives you 0.013

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4 years ago

What is the average rate of change of f over the interval -1 on a separate sheet of paper and upload the work as a photo. Show a

Anarel [89]

Answer:

Average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

Step-by-step explanation:

We need to find average rate of change of f over the interval -1 < x < 1

The function given is: $f(x)= x^2-x-1$

The formula used to find average rate of change is:

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}$

We have, a = -1 and b = 1

Finding f(b) when b=1

$f(x)=x^2-x-1\\f(1)=(1)^2-(1)-1\\f(1)=1-1-1\\f(1)=-1$

Now, finding f(a), when a= -1

$f(x)=x^2-x-1\\f(-1)=(-1)^2-(-1)-1\\f(1)=1+1-1\\f(-1)=2-1\\f(-1)=1$

Now, putting values and finding average rate of change

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}\\Average\:rate\:of\:change=\frac{f(1)-f(-1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-(1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-1}{1+1}\\Average\:rate\:of\:change=\frac{-2}{2}\\Average\:rate\:of\:change=-1$

So, average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

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3 years ago