1. What is the image of A(-4, 1) using the translation (x, y) --> (x + 6, y

About A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows: A: The sum on the two

Flauer [41]

$p(A) = \frac{1}{2}$

$p(B) = \frac{1}{6}$

$p(C) = \frac{1}{6}$

$P(A | C)=\frac{1}{2}$

Solution:

The probability of an event is given as:

$\text { probability of an event }=\frac{\text { number of favorable outcomes }}{\text { total number of outcomes }}$

In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}

First let us calculate p(A):

The event is defined as: The sum on the two dice is even

Sum on two dice is even if and only if either both dice turn up odd or both even.

The odd outcomes in thowing a single die = 3 {1, 3, 5}

The even outcomes in throwing a single die = 3 {2, 4, 6}

The probability that both turn up odd is:

$\text { probability of both die turing up odd }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

Similarly, the probability that both turn up even is:

$\text { probability of both die turing up even }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}$

probability that the sum on two dice is even = probability that both turn up odd + probability that both turn up even

$\text { probability of sum on two dice is even }=\mathrm{p}(\mathrm{A})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Thus $p(A) = \frac{1}{2}$

Let us calculate p(B):

The event B is defined as: The sum on the two dice is at least 10

The total possible outcomes of two die is given as:

{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :

atleast 10 means that sum can be 10 or greater than 10

{(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)}

Here favourable outcomes = 6

Total number of outcomes = 36

Hence, the probability that the sum of the two dice will be at least 10 is:

$\text { probability that the sum of the two dice will be at least } 10=\frac{6}{36}=\frac{1}{6}$

Thus $p(B) = \frac{1}{6}$

Let us calculate p(C):

The event C is defined as: The red die comes up 5

Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

$\text { probability of red die comes up } 5 \text { is the event }=\frac{6}{36}=\frac{1}{6}$

Thus $p(C) = \frac{1}{6}$

B) What is p(A l C)

$P(A | C)=\frac{p(A \cap C)}{P(C)}$

$\mathrm{A} \cap \mathrm{C}=\{(1,5),(3,5),(5,5)\}$

$p(A \cap C)=\frac{3}{36}=\frac{1}{12}$

$P(A | C)=\frac{p(A \cap C)}{P(C)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}$

Thus $P(A | C)=\frac{1}{2}$

8 0

3 years ago

Can someone solve -3 = x + 7 - 4?

Blababa [14]

Answer:

x= -6

v= 1

v= 2

Step-by-step explanation:

-3= x+7-4, combine like terms by subtracting seven from four

-3= x+3, subtract three from both sides

-6= x

8v-3+3v= 8, combine like terms by adding 8v and 3v

11v-3= 8, add three to both sides

11v= 11, divide both sides by 11

v=1

-v+3v=4, combine like terms by adding -v and -3v

2v= 4, divide both sides by 2

v= 2

3 0

3 years ago

Read 2 more answers

The Leukemia and Lymphoma Society sponsors a 5K race to raise money. It receives $55 per race entry and $10,000 in donations, bu

loris [4]

Donations = $10,000
Revenue = $55 per race enry
Expense = $15 per race entry
Budget ≥ $55,000

Let x = the number of race entries.
Total revenue = Donations + Race entries - Expenses
Therefore
10000 + 55x - 15x ≥ 55000
40x ≥ 45000
x ≥ 1125

Answer: Race entries ≥ 1125

6 0

4 years ago

What's one minus three eighths?

Wewaii [24]

Five eighths l, think about it as 8 eights minus 3 8ths. Keep the bottom subtract the top

8 0

3 years ago

Read 2 more answers

Help me with the answer please

Andreyy89

Brainly bot here, telling you that your account has been temporarily banned, and the only people that can answer your questions are the other banned accounts. dm the managers to find out more information.

6 0

3 years ago

1. What is the image of A(-4, 1) using the translation (x, y) --> (x + 6, y - 3)?