Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.
For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.
<h3>Binomial probability distribution
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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:
- One in ten students are named Joe, hence
.
- There are 15 students in the class, hence
.
The probability that at least one of them is named Joe is:

In which:


Then:

0.7941 = 79.41% probability that at least one of them is named Joe.
To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377
Answer:
F(x) = 3x^2 + 1, G(x) = 2x - 3, H(x) = x
F(3) = 3(3)^2 + 1 = 27 + 1 = 28
G(4) = 2(4) - 3 = 8 - 3 = 5
2H(5) = 2(5) = 10
F(3) + G(4) - 2H(5) = 28 + 5 - 10
F(3) + G(4) - 2H(5) = 23
Step-by-step explanation:
I believe I'm not sure tho
Answer:
y=-2x+2
Step-by-step explanation:
If you bring the 4x to the other side by subtracting it, you will have 2y=-4x+8
then divide by the 2 to get y=-2x=4
then you substitute the 5 in for y and the -1 in for x. multiply and then subtract
12 times 2 + 9 times 2 = 24 + 18 = 42 divided by 16 = 2.625
Answer: 3/4 I’m pretty sure. I looked at it and got confused so not 100% sure.