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Kaylis [27]
3 years ago
8

If you were to flip a coin twenty times, approximately how many times do you think heads would come up?

Mathematics
2 answers:
Gemiola [76]3 years ago
8 0
Theoretically 10 because the 50/50 chance realistically though you'll get more or less
Llana [10]3 years ago
7 0
The probability of heads on any toss is just as likely as the probability of tails because each coin toss is an independent event.  In other words two events related in such a way that knowing about the occurrence of one event has no effect on the probability of the other event, in others words independent.

It does not matter if you have flipped a coin 5 times and it was heads, it is not "due" to come up tails.  This is an incorrect assumption just like it is incorrect to believe you are "on a roll" because 5 times it was heads.  
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Is the relation a function? If so why or why not? {(-3,2),(-1,0),(1,0),(5,-2)}
Anastasy [175]

Answer:

It's a function

Step-by-step explanation:

Functions can only have one output for each input. And since all the x values for each point are different, there is only one output (y-value) for each input(x-value).

8 0
3 years ago
87+56=____<br> (f r e e p o i n t s)<br> prolly gonna give brainliest
pogonyaev

Answer: 143

Step-by-step explanation:

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5 0
3 years ago
PQ and RS are two lines that intersect at point T, as shown below:
Setler79 [48]
Our approach to answering this question is to eliminate the choices until we are left with only one. 

(1) FALSE. The given figures are lines and can extend indefinitely.
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3 years ago
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sergey [27]
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7 0
3 years ago
Read 2 more answers
In order to determine the effectiveness of a new measles antibody test, it is administered to 500 people chosen at random. The n
myrzilka [38]

Answer:

The probability that a randomly chosen person has measles antibodies in his/her blood if the new test is positive = 0.9796

Step-by-step explanation:

The event that someone tests positive = P(T)

The event that someone has antibodies = P(A)

The event that someone does not have antibodies = P(A')

The new test was positive when administered to 96% of those who have the antibodies.

This probability = P(T n A) = 0.96

The new test gave positive results in 2% of those who do not have them.

This probability = P(T n A') = 0.02

The probability that a randomly chosen person has measles antibodies in his/her blood if the new test is positive = P(A|T)

This conditional probability is given as

P(A|T) = P(T n A) ÷ P(T)

P(T) is given as

P(T) = P(T n A) + P(T n A') = 0.96 + 0.02 = 0.98

P(A|T) = P(T n A) ÷ P(T) = 0.96 ÷ 0.98 = 0.9796

Hope this Helps!!!

5 0
3 years ago
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