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3 years ago

A(n)=-4-2(n-1) find the 12th term in the sequence.

Mathematics

1 answer:

Art [367]3 years ago

4 0

Substitute n = 12:

$a(n)=-4-2(n-1)\\\\a(12)=-4-2(12-1)=-4-2(11)=-4-22=-26$

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2 years ago

A track star runs two races on a certain day. The probability thathe wins the first race is 0.7, the probability that he wins th

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Answer:

a) 80% probability that he wins at least one race.

b) 30% probability that he wins exactly one race.

c) 20% probability that he wins neither race.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that he wins the first race.

B is the probability that he wins the second race.

C is the probability that he does not win any of these races.

We have that:

$A = a + (A \cap B)$

In which a is the probability that he wins the first race but not the second and $A \cap B$ is the probability that he wins both these races.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that he wins both races is 0.5.

This means that $A \cap B = 0.5$

The probability that he wins the second race is 0.6

This means that $B = 0.6$

$B = b + (A \cap B)$

$0.6 = b + 0.5$

$b = 0.1$

The probability that he wins the first race is 0.7.

This means that $A = 0.6$

$A = a + (A \cap B)$

$0.7 = a + 0.5$

$a = 0.2$

A) he wins at least one race.

This is

$P = a + b + (A \cap B) = 0.2 + 0.1 + 0.5 = 0.8$

There is an 80% probability that he wins at least one race.

B) he wins exactly one race.

This is

$P = a + b = 0.2 + 0.1 = 0.3$

There is a 30% probability that he wins exactly one race.

C) he wins neither race

Either he wins at least one race, or he wins neither. The sum of these probabilities is 100%.

From a), we have that there is an 80% probability that he wins at least one race.

So there is a 100-80 = 20% probability that he wins neither race.

6 0

3 years ago