Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.
<h3>What is the binomial distribution formula?</h3>
The formula is:


The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 0.54% of the calls receive a busy signal, hence p = 0.0054.
- A sample of 1300 callers is taken, hence n = 1300.
The probability that at least 5 received a busy signal is given by:

In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:






Then:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

0.8295 = 82.95% probability that at least 5 received a busy signal.
More can be learned about the binomial distribution at brainly.com/question/24863377
#SPJ1
1 4 . 8 3333
6 8 9 50
14.8333
Step-by-step explanation:
step 1. if k is a zero then (x - 1) is a factor of 4x^3 + 23x^2 - 34x + 7.
step 2. we "can" use long division but synthetic division is easier!
step 3. 1 4 23 -34 7
4. 4. 27. -7.
4 27. -7. 0 (must get 0)
step 4. so the factors are (x - 1)(4x^2 + 27x - 7).
step 5. please note if you multiply these factors together you will get the original polynomial!
Step-by-step explanation:
