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3 years ago

Stacy rolls a pair of six-sided fair dice.

Mathematics

1 answer:

Nutka1998 [239]3 years ago

5 0

Answer:

The probability that the sum of the numbers rolled is either a multiple of 3 or an even number is 24/36 = 2/3

The two events are mutually exclusive

Step-by-step explanation:

we first need to create a table of the sample space from the experiment, the sum of the numbers rolled. Find the attached for a depiction of the possible outcomes.

The probability that the sum of the numbers rolled is either a multiple of 3 or an even number will be given by counting the numbers that are either even or multiples of 3 and then dividing by the number of possible outcomes, 36.

In our case this will be;

24/36 = 2/3

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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

On a camping trip with friends, you read about mosquito activity after dark. The number of mosquitos in your campground can be m

Dmitrij [34]

Answer:

The answer is 54

Step-by-step explanation:

If you follow PEMDAS, and x = 3, then you get 54. As an example, x4 is 12. You change X to 3, then multiply (3)4=12. If you follow the rest of the equation, you get 54.

6 0

3 years ago

Read 2 more answers

Is the following shape a square? How do you know?

Troyanec [42]

Answer:

Step-by-step explanation:

it is a rectangle. each opposite sides ARE parallel but not every side is equal like a square so B

8 0

3 years ago

How can you find a place value in dicimals?

Verizon [17]

Well 1234567 is a number and
7 is in one's place
6 is in ten's place
5 is in hundred's place
4 is in thousand's place
3 is in ten thousand's place
2 is in thousand's place
1 is in million's place

I hope I have answered it right

5 0

3 years ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

<u>Step-by-step explanation:</u>

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

3 0

3 years ago