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katrin [286]
3 years ago
6

A speedy river barge bound for New Orleans, Louisiana leaves Baton Rouge at 9:00 a.m. and travels at a speed of 10 miles per hou

r. A rail transport freight also bound for New Orleans leaves Baton Rouge at 1:30 p.m. on the same day. The train travels at 25 miles per hour, both the barge and train will travel 100 miles to reach New Orleans. How far will the train travel before catching up to the barge?
Mathematics
2 answers:
zhuklara [117]3 years ago
8 0

Answer:

Step-by-step explanation:

A rail transport freight train also bound for New Orleans leaves Baton Rouge at 1;30 PM on the same day. The train travels at 25 MPH and the river barge travels at 10 MPH. ... the barge leaves at 9:00 am so that would be at 9 hours past midnight. the train leaves at 1:30 pm so that would be at 13.5 hours past midnight.

kirza4 [7]3 years ago
4 0
Okay i get it now
1) At 1.30 p.m. the barge has already traveled 4.5 X 10 = 45 miles.

To catch up it will take the train 25-10 = 15miles

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3 years ago
According to a study done by a university​ student, the probability a randomly selected individual will not cover his or her mou
Sergio [31]

Using the binomial distribution, it is found that:

a) There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when​ sneezing.

b) There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when​ sneezing.

c) 9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when​ sneezing.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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.

The values of the parameters are given as follows:

n = 18, p = 0.267.

Item a:

The probability is P(X = 6), hence:

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

P(X = 6) = C_{18,6}.(0.267)^{6}.(0.733)^{12} = 0.1618

There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when​ sneezing.

Item b:

The probability is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).

Then:

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

P(X = 0) = C_{18,0}.(0.267)^{0}.(0.733)^{18} = 0.0037

P(X = 1) = C_{18,1}.(0.267)^{1}.(0.733)^{17} = 0.0245

P(X = 2) = C_{18,2}.(0.267)^{2}.(0.733)^{16} = 0.0758

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0037 + 0.0245 + 0.0758 = 0.104.

There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when​ sneezing.

item c:

We have to look at the mean and the standard deviation, given, respectively, by:

  • E(X) = np = 18 x 0.267 = 4.81.
  • \sqrt{V(X)} = \sqrt{18(0.267)(0.733)} = 1.88

9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when​ sneezing.

More can be learned about the binomial distribution at brainly.com/question/24863377

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