Https://mathbitsnotebook.com/Geometry/Circles/CRAngles.html
Answer:
Volume equals 56.6 when rounded to the nearest tenth
Step-by-step explanation:
YOU NEED A CALCULATOR TO SOLVE THIS!
To find the volume of a cylinder you need the formula
Now that you have the formula you plug in the radius and the height

Then solve



Answer:
There was a 1.93% probability of having multiples (twins or more) in 1980.
Step-by-step explanation:
There were 3,612,258 births in the US in 1980.
Of those,
68339 + 1337 = 69676 were multiplies.
So
What was the probability of having multiples (twins or more) in 1980?

There was a 1.93% probability of having multiples (twins or more) in 1980.
Answer:
One convergence criteria that is useful here is that, if aₙ is the n-th term of this sequence, then we must have:
Iaₙ₊₁I < IaₙI
This means that the absolute value of the terms must decrease as n increases.
Then we must have:

We can write this as:

If we assume that n is a really big number, then:
n + 1 ≈ 1
And we can write:

Then we have the inequality

And remember that this must be in absolute value, then we will have that:
-1 < (x - 2)/3 < 1
-3 < x - 2 < 3
-3 + 2 < x < 3 + 2
-1 < x < 5
The first option looks like this, but it uses the symbols ≤≥, so it is not the same as this, then the correct option will be the second.