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givi [52]
3 years ago
13

You roll two dice. What is the probability that the sum of the dice is prime and both dice show the number 6?

Mathematics
1 answer:
Eduardwww [97]3 years ago
7 0
 <span>The probability of both dice showing 6: 

The total possible outcomes of rolling 2 dice is 36. (1+1,1+2,..., 6+5, 6+6). Only one of these 36 rolls give two 6s. Hence, The probability of both dice showing 6 is 1/36 (1 in 36). 


We have 1 + 2 + 4+ 6 +2 ways to get prime rolls, which is a total of 15 ways. 
Hence the probability of getting a prime is 15/36.</span>
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Simplify (1-√3)(⅓+√3) leaving your answer in the form p+q√3​
Mars2501 [29]

Answer:

\frac{-8 +2\sqrt{3} }{3}

Step-by-step explanation:

When working with surds we need to take note of the roots present there.

To expand this equation we can do it the following way noting that √3 X √3 = 3

<em></em>

<em>Expanding (1-√3)(⅓+√3)</em>

1 X 1/3 = 1/3

1 X √3 = √3

-√3 X 1/3 =-√3/3

√3 X √3 = 3

hence, expanding the equation, we have

1/3 + √3 -√3/3 + 3

We can simply group the like terms and add them up.

[1/3 +3] +[√3-√3/3]

10/3 + \frac{2\sqrt{3} }{3}

= \frac{-8 +2\sqrt{3} }{3}

8 0
3 years ago
What is 8 - 7 1/4 i need the answer fast
Citrus2011 [14]

Answer:

3/4

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and con
sladkih [1.3K]

65 sequences.

Lets solve the problem,

The last term is 0.

To form the first 18 terms, we must combine the following two sequences:

0-1 and 0-1-1

Any combination of these two sequences will yield a valid case in which no two 0's and no three 1's are adjacent

So we will combine identical 2-term sequences with identical 3-term sequences to yield a total of 18 terms, we get:

2x + 3y = 18

Case 1: x=9 and y=0

Number of ways to arrange 9 identical 2-term sequences = 1

Case 2: x=6 and y=2

Number of ways to arrange 6 identical 2-term sequences and 2 identical 3-term sequences =8!6!2!=28=8!6!2!=28

Case 3: x=3 and y=4

Number of ways to arrange 3 identical 2-term sequences and 4 identical 3-term sequences =7!3!4!=35=7!3!4!=35

Case 4: x=0 and y=6

Number of ways to arrange 6 identical 3-term sequences = 1

Total ways = Case 1 + Case 2 + Case 3 + Case 4 = 1 + 28 + 35 + 1 = 65

Hence the number of sequences are 65.

Learn more about Sequences on:

brainly.com/question/12246947

#SPJ4

8 0
2 years ago
Is 5.3 repeating rational
Goshia [24]

Repeating decimals are considered rational numbers because they can be represented as a ratio of two integers. If a number is terminating or repeating, it must be rational if a decimal is both non terminating and non repeating, the number is irrational.

So yes.

4 0
4 years ago
Andy made 7 baskets during a basketball game each basket was worth 2 or 3 points how many different numbers could represent the
Phantasy [73]

Answer:

2+3+2+3+2+3+2=17 because each was worth 2 or 3

5 0
2 years ago
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