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Lesechka [4]
3 years ago
6

Brian rolls a number cube twice. What is the probability that the sum of the 2 rolls is less than 6 given that the first roll is

a 3?
Mathematics
2 answers:
Alona [7]3 years ago
8 0
If the first roll is a three, then the only chance of the sum of the two numbers being less than six is if you roll a 1 or a 2.
That is 2/6 chance, or simplified, a 1/3 chance.

I hope this Helps!
den301095 [7]3 years ago
7 0

Answer:  The required probability is \dfrac{1}{3}.

Step-by-step explanation: Given that Brian rolls a number cube twice.

We are to find the probability that the sum of the 2 rolls is less than 6 given that the first roll is a 3.

Let, 'S' be the sample space for the experiment of rolling the second number cube when the first roll is a 3.

Then, S = {(3, 1), (3, 2), (3, 3), (3,4), (3, 5), (3, 6)}   ⇒  n(S) = 6.

And, let 'E' be the event that the sum of the two rolls is less than 6.

Then, E = {(3, 1), (3, 2)}  ⇒  n(E) = 2.

Therefore, the probability that the sum of the 2 rolls is less than 6 given that the first roll is a 3 is

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{2}{6}=\dfrac{1}{3}.

Thus, the required probability is \dfrac{1}{3}.

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3. .5*60 because .5 represents the 5% and "of" stands for multiplication
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What is the equivalent decimal to this fraction? 10/45 0.2 0.222222222... 0.232323232323... 0.181818181818...​
Ivan

Factorize the numerator and denominator, then simplify:

10/45 = (2×5)/(9×5) = 2/9

Now,

1/9 = 1/10 + 1/90

1/90 = 1/10 × 1/9 = 1/10 × (1/10 + 1/90) = 1/100 + 1/900

1/900 = 1/100 × 1/9 = 1/100 × (1/10 + 1/90) = 1/1000 + 1/9000

and so on, which is to say

1/9 = 1/10 + 1/100 + 1/1000 + …

or

1/9 = 0.111…

so that multiplying both sides by 2 gives

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5 0
2 years ago
Which one of these is a square number? <br><br> 15<br> 39<br> 81<br> 132
coldgirl [10]

81 = 9^2

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3 years ago
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Three people go to a restaurant. Their bill comes to $78.00. They decide to split the cost. One person pays $4.50; the next pers
Nikitich [7]

Answer:

$60.00

Step-by-step explanation:

First person:

$4.50

Second person:

3 times $4.50 = $13.50

Total of first two people:

$4.50 + $13.50 = $18.00

Third person:

$78.00 - $18.00 = $60.00

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8 0
3 years ago
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Suppose that two openings on an appellate court bench are to be filled from current municipal court judges. The municipal court
Ksju [112]

Answer:

(a)\dfrac{92}{117}

(b)\dfrac{8}{39}

(c)\dfrac{25}{117}

Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

  • A man is appointed first and a woman is appointed next.
  • A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed)=P(MW)+P(WM)

=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}

(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

  • A man is appointed first and a woman is appointed next.
  • A woman is appointed first and a man is appointed next.
  • Two women are appointed

P(at least one woman is appointed)=P(MW)+P(WM)+P(WW)

P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}

In Part B, P(MW)+P(WM)=\frac{8}{39}

Therefore:

P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}

5 0
3 years ago
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