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Virty [35]
3 years ago
5

!!!! URGENT evaluate the geometric series below !

Mathematics
1 answer:
True [87]3 years ago
6 0

Answer:

C

Step-by-step explanation:

S(6)= -2(1-(-3)^6)/1-(-3)

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Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete
icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question:

Fleet of 17 means that N = 17

4 are carrying nucleas weapons, which means that k = 4

9 are destroyed, which means that n = 9

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

P(X > 1) = 1 - P(X \leq 1)

In which

P(X \leq 1) = P(X = 0) + P(X = 1)

So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294

P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0
2 years ago
What is the net force on a 1,000 kg object accelerating at 3 m/s2 ?
anzhelika [568]
33 \leq m+1.4
6 0
3 years ago
In which year will 67% of babies be born out of wedlock?
Leviafan [203]

Based on the trend of the increase in children born out of wedlock, if this trend keeps increasing, the year with 67% of babies born out of wedlock will be 2055 .

<h3>What year will 67% of babies be born to unmarried parents?</h3><h3 />

In 1990, the 28% of children were born out of wedlock and this trend was increasing by 0.6% per year.

If the trend continues, the number of years till 67% of children born out of wedlock will be:

= (67%  - 28%) / 0.6%

= 65 years

The year will be:

= 1990 + 65

= 2055

The first part of the question is:

According to the National Center for Health Statistics, in 1990, 28% of babies in the United States were born to parents who were not married. Throughout the 1990s, this percentage increased by approximately 0.6 per year.

Find out more on benefits of marriage at brainly.com/question/12132551.

#SPJ1

3 0
1 year ago
7. After boiling, the 18 liters of water was reduced to 7 liters. How many water was
e-lub [12.9K]

Answer:

C

Step-by-step explanation:

Sorry if its wrong bro

4 0
2 years ago
What is the solutiion to this system 3x+2y=8 -x=+y=-11
mr Goodwill [35]

Answer:

2, 19, -3

Step-by-step explanation:

3 0
3 years ago
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