Probability(error in one block)= 1/5
<span>Probability(error in 3 blocks)= 3*(1/5)= 0.6</span>
Let,
0 = No error
<span> 1 = error
</span>Hence.
E(x) = 0.4 x 0 x+0.6 + 1
= 1.6
Answer:
-384x - 120y
Step-by-step explanation:
General Concepts and Formulas:
Brackets
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
Steps:
1) Distribute

We must distribute the expression above by multiplying -12 by the coefficients and then adding the variables.
So first, multiple -12 by 32 and add the variable 'x' when getting the product:
-12(-384x - 10y)
Now that we have one part, let's solve the other part.
We now must multiple -12 by -10, which gets us 120, now we must plug in:
(-384x - 120y)
(Remove parenthesis)
⇒ -384x - 120y
Answer:
S = 8
Step-by-step explanation:
An infinite geometric series is defined as limit of partial sum of geometric sequences. Therefore, to find the infinite sum, we have to find the partial sum first then input limit approaches infinity.
However, fortunately, the infinite geometric series has already set up for you. It’s got the formula for itself which is:

We can also write in summation notation rather S-term as:

Keep in mind that these only work for when |r| < 1 or else it will diverge.
Also, how fortunately, the given summation fits the formula pattern so we do not have to do anything but simply apply the formula in.

Therefore, the sum will converge to 8.
Please let me know if you have any questions!
I don't know how to make a 'greater than' sign.
6x is greater than x + 20
6x - x is greater than 20
5x is greater than 20
x is greater than 20 divided by 5
x is greater than 4
So, any number greater than 4 will work in this equation.