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Ulleksa [173]
3 years ago
10

2020 people formed a line. In one move, you are allowed to exchange the positions of two people, separated by one other person.

Can you always sort the people form tallest to shortest using these moves?
Mathematics
1 answer:
Vladimir79 [104]3 years ago
7 0

Answer:

yes you CAN you can use the descending to ascending formular but on people in which they might be a mixup of tallest and shortest

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Answer: C 1/2mL LMN= mL LMO

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The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.
mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.96516 probability of a chipmunk living through the year, thus p = 0.96516

Item a:

  • Two is P(X = 2) when n = 2, thus:

P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

  • Six is P(X = 6) when n = 6, then:

P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

  • At least one not living is:

P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

6 0
3 years ago
A car is travelling at exactly 6 miles per hour and accelerates at a constant rate to exactly 65 miles per hour.
vichka [17]

Answer:

14878.04878miles/hours^2

Step-by-step explanation:

Let's find a solution by understanding the following:

The acceleration rate is defined as the change of velocity within a time interval, which can be written as:

A=(Vf-Vi)/T where:

A=acceleration rate

Vf=final velocity

Vi=initial velocity

T=time required for passing from Vi to Vf.

Using the problem's data we have:

Vf=65miles/hour

Vi=6miles/hour

T=14.8seconds

Using the acceleration rate equation we have:

A=(65miles/hour - 6miles/hour)/14.8seconds, but look that velocities use 'hours' unit while 'T' uses 'seconds'.

So we need to transform 14.8seconds into Xhours, as follows:

X=(14.8seconds)*(1hours/60minutes)*(1minute/60seconds)

X=0.0041hours

Using X=0.0041hours in the previous equation instead of 14.8seconds we  have:

A=(65miles/hour - 6miles/hour)/0.0041hours

A=(61miles/hour)/0.0041hours

A=(61miles)/(hour*0.0041hours)

A=61miles/0.0041hours^2

A=14878.04878miles/hours^2

In conclusion, the acceleration rate is 14878.04878miles/hours^2

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