Using the binomial distribution, it is found that the probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.
For each person, there are only two possible outcomes, either they need correction for their eyesight, or they do not. The probability of a person needing correction is independent of any other person, hence, the binomial distribution is used to solve this question.
<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:
- A survey showed that 77% of us need correction, hence p = 0.77.
- 13 adults are randomly selected, hence n = 13.
The probability that at least 12 of them need correction for their eyesight is given by:

In which:



Then:

The probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.
More can be learned about the binomial distribution at brainly.com/question/24863377
Answer:
p = 4
Step-by-step explanation:
Given equation:

<u>Standard equation of a circle:</u>

(where
is the centre of the circle, and
is the radius)
If you expand this equation, you will see that the coefficient of
is always one.
Therefore, 

<u>Additional information</u>
To rewrite the given equation in the standard form:





So this is a circle with centre (2, -3) and radius √29
The answer I got is 178.72. U divided $8,936 by 2%
Since h represents the height of the ball at any given time, t, let h = 25, such that the ball will be 25m high at t.
Now, we have 25 = 20t - 5t²
5t² - 20t + 25 = 0
t² - 4t + 5 = 0

Since the discriminant is less than zero, there are no solutions.
Hence, the ball will never be 25m high.