5 - 3(x+2) for math (algebra)
2 answers:
You might be interested in
<h3>Answer:</h3>
In this question, it's asking you to find how much percentage the circle graph is for "A" papers.
To solve this question, we would need to use information from the question.
Important information:
- Graded 50 English research papers
- 12 of those papers had an "A" grade
With the information above, we can solve the question.
We know that there are 12 research papers that received an A and there are 50 research papers in total.
We would divide 12 by 50 in order to find the percentage of the papers that got an A.
When you divide, you should get 24.
This means that 24% of the circle graph is devoted to "A" papers.
<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>Answer:
(a) The value of P (None) is 0.062.
(b) The value of P(at least one) is 0.938.
(c) The value of P(at most one) is 0.253.
(d) The event is not unusual.
Step-by-step explanation:
Let <em>X</em> = number of households watching the show.
The probability of the random variable <em>x</em> is, P (X) = <em>p</em> = 0.18.
The sample selected for the survey is of size, <em>n</em> = 14
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 14 and <em>p</em> = 0.18.
The probability of a Binomial distribution is computed using the formula:
(a)
Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:
Thus, the value of P (None) is 0.062.
(b)
Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
Thus, the value of P(at least one) is 0.938.
(c)
Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)
Thus, the value of P(at most one) is 0.253.
(d)
An event that has a very low probability of occurrence is known as an unusual event.
The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.
This probability value is not low.
Hence, the event is not unusual.