I have the answer but brainly is not working
Answer:
-2n²
Step-by-step explanation:
Using the image attached:
We see that the second differences (blue ones) are all equal so we conclude that this is a quadratic sequence.
The quadratic sequence has the form:

To find the value of a we just divide second difference ( -4 ) by 2:
Now we have:
Substitute
and
into the equation above:
Since
= -2 and
= -8 , after simplification we have:
The solution of this system is:
The general term is :

Answer:
The probability is 0.971032
Step-by-step explanation:
The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.
The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:
(eq. 1)
So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:
P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)
We can also calculated that as:
P(x ≥ 5) = 1 - P(x ≤ 4)
Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)
Then, if we calculate every probability using eq. 1, we get:
P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765
P(x ≤ 4) = 0.028968
Finally, P(x ≥ 5) is:
P(x ≥ 5) = 1 - 0.028968
P(x ≥ 5) = 0.971032
Answer:
the original price was $46.12
Step-by-step explanation:
if you add 5 back to that number.
Answer: 3/2
Step-by-step explanation:
<u>Given</u>
6/10 + 9/10
<u>Add the numerator together</u>
= (6 + 9) / 10
<u>Simplify by addition</u>
= 15 / 10
<u>Simplify to simplest form by division</u>
= (15 ÷ 5) / (10 ÷ 5)
= 3 / 2
Hope this helps!! :)
Please let me know if you have any questions