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Misha Larkins [42]
3 years ago
6

PLEASE ANSWER, I AM STRUGGLING! WILL GIVE POINTS! <3

Mathematics
2 answers:
Oliga [24]3 years ago
7 0
X/2 = 68.2
Given that Colin’s number is X, he divided it by 2. Hence it will be x/2 =68.2. Hope this helps! :)
prisoha [69]3 years ago
4 0
Here is the working
Hope it helps !!

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6x-8=16
6x-8+8=16+8
6x=24
6x=24
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6 6
X= 4
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16% of what number is 14 ? <br><br><br><br> ......
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Step-by-step explanation:

x=100%

14=16%

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Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let A be the event that th
Blababa [14]

Answer:

- P(A) = (1/2) = 0.5

- P(B) = (1/2) = 0.5

- P(A n B) = (1/4) = 0.25

The two events, A and B, are independent of each other.

Step-by-step explanation:

A fair six sided die has 6 numbers with 3 even and 3 odd numbers.

A fair four sides die has 4 numbers with 2 even and 2 odd numbers.

A is the event that the number on the six-sided die is an even number.

B is the event that that number on the four-sided die is an odd number.

Note that the probability of an event, E is the number of elements in that event divided by the total numner of events in the sample space.

P(E) = n(E) ÷ n(S)

a) What is P(A), the probability that the six-sided die is an even number?

P(A) = n(A) ÷ n(S)

n(A) = 3

n(S) = 6

P(A) = (3/6) = (1/2) = 0.5

b) What is P(B), the probability that the four-sided die is an odd number?

P(B) = n(B) ÷ n(S)

n(B) = 2

n(S) = 4

P(A) = (2/4) = (1/2) = 0.5

c) What is P(A and B), the probability that the six-sided die is an even number and the four-sided die is an odd number?

The sample space for this includes

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

n(S) = 24

(A n B) = (2,1), (2,3), (4,1), (4,3), (6,1), (6,3)

n(A n B) = 6

P(A n B) = (6/24) = (1/4) = 0.25

d) Are events A and B independent?

Two events are said to be independent of each other if the chances of one happening does not depend on the chances of the other happening and vice versa.

Mathematically, independent events are tested thus

P(A n B) = P(A) × P(B)

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = 0.25, P(A) = P(B) = 0.5

0.25 = 0.5 × 0.5

0.25 = 0.25 (hence, the two events are truly independent)

P(A|B) = P(A n B) ÷ P(B)

P(A n B) = 0.25, P(B) = 0.5

P(A|B) = (0.25/0.5) = 0.5 = P(A) (another proof of their independence)

P(B|A) = P(A n B) ÷ P(A)

P(A n B) = 0.25, P(A) = 0.5

P(B|A) = (0.25/0.5) = 0.5 = P(B) (further proof of their independence)

Hope this Helps!!!

3 0
3 years ago
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