Help?? Due tomorrow!!

Assume that there are an equal number of births in each month so that the probability is that a person chosen at random was born

NikAS [45]

Answer:

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of a person being in May:

May has 31 days in a year of 365. So

$p = \frac{31}{365} = 0.0849$

Group of 20 friends:

This means that $n = 20$

What is the probability that at the May celebration, exactly two members of the group have May birthdays?

This is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{20,2}.(0.0849)^{2}.(0.9151)^{18} = 0.2773$

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

3 0

3 years ago

i have 6 digits. One of my 3s is worth 300.000. The other is worth 1/10 as much. My 6 is worth 600.The rest of my digits are zer

motikmotik

The way you wrote the problem makes the answer 330,600. I think your decimals were meant to be commas. If I am wrong, please forgive me.

3 0

3 years ago

Four students are called to sit in four chairs on

bekas [8.4K]

There is a 50/50 chance Bob and Wanda will sit next to each other.

8 0

4 years ago

Identify the data set that could be quadratic. HELP ASAP!!

Ne4ueva [31]

Answer:

Option B can be quadratic.

Step-by-step explanation:

Take a look at the data.

A quadratic equation is of the form ,

$y = ax^{2} +bx + c$ ,

where, a,b,c are constants.

To find an unknown equation of 3 variables,

we need 3 points lying on the equation.

here they have given, 5 points, meaning all should lie on the curve.

For first option, inserting first 3 points to find equation, we get equation as,

$y = x^{2} -4x +11$ , but the rest points don't satisfy the curve.

So first option is not quadratic.

Similarly, it can be shown that option C and D are also not quadratic.

While, in option B, it is clear that for every y, x is y squared,

$x = y^{2}$ , thus quadratic in nature.

6 0

3 years ago

Answer the problem :)

Tomtit [17]

The domain is the value of x. In this case, -3≤x≤7
the range is the value of y. in this case, -1≤y≤9

this is not a function, because the same x value has two corresponding y values. For example, when x=5, y=0 or y=8

It IS a function if every x value has only one corresponding y value.

5 0

3 years ago