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kari74 [83]
4 years ago
15

11 ounces to 5 ounces

Mathematics
2 answers:
Maurinko [17]4 years ago
6 0

Answer:

(11/5) (a ratio)

Step-by-step explanation:

11 ounces to 5 ounces could be rewritten as

 11 oz.

---------- = (11/5) (a ratio)

 5 oz

harkovskaia [24]4 years ago
4 0

Answer: what is the question to this?

Step-by-step explanation: thanks let me know okay

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24/2 + (-2x5) = 12 + (-10) = 2
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Step-by-step explanation:

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What is the volume of a sphere with a radius of 18 units?
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We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely. (a) Given that the roll results
boyakko [2]

Answer:

a) \frac{1}{7}

b) \frac{2}{15}

Step-by-step explanation:

Given : We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely.

The outcomes are :

(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)

(a) Given that the roll results in a sum of 3 or more, find the conditional probability that doubles (the first and the second rolls result in the same number) are rolled.

Let P(A) be the probability of event that the roll results in a sum of 3 or more.

Except (1,1) rest sum is 3 or greater than 3.

So, P(A)=\frac{35}{36}

Now, P(A and B) is that the roll results in a sum of 3 or more and that doubles (the first and the second rolls result in the same number) are rolled.

i.e. (2,2), (3,3), (4,4), (5,5), (6,6) - 5

P(A\cap B)=\frac{5}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{5}{36}}{\frac{35}{36}}

P(B|A)=\frac{5}{35}

P(B|A)=\frac{1}{7}

(b) Given that the two dice land on different numbers, find the conditional probability that the sum is 5.

Let P(A) be the probability of event that two dice land on different numbers.

Except (1,1),(2,2), (3,3), (4,4), (5,5), (6,6) rest two dice land on different numbers.

So, P(A)=\frac{30}{36}

Now, P(A and B) is that two dice land on different numbers and the sum is 5.

i.e. (1,4), (2,3), (3,2), (4,1) - 4

P(A\cap B)=\frac{4}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{4}{36}}{\frac{30}{36}}

P(B|A)=\frac{4}{30}

P(B|A)=\frac{2}{15}

4 0
3 years ago
Can somebody help me please.
Fiesta28 [93]
I’m pretty sure it is 57?
7 0
3 years ago
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