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AfilCa [17]
3 years ago
15

The graph below shows the number of visitors at a theme park. How many more visitors in total went to the park in June, July and

August in Year 2 than Year 1?

Mathematics
1 answer:
Olin [163]3 years ago
8 0
From the information given by the graph

Year 1 visitors
June = 9450
July = 9500
August = 9600

Year 2 visitors
June = 9450
July = 9600
August = 9800

Total visitor Year 1 = 9450 + 9500 + 9600 = 28550
Total visitor Year 2 = 9450 + 9600 + 9800 = 28850

Visitor in Year 2 is 28850-28550 = 300 more than visitor in Year 1
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Step-by-step explanation:

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The high temperature for one town last year was 79°F The low temperature for the same town was - 3°F What was the difference bet
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I have a F in this class may someone help me..?
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Naval intelligence reports that 99 enemy vessels in a fleet of 1818 are carrying nuclear weapons. If 99 vessels are randomly tar
oee [108]

Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, 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 18 means that N = 18

9 are carrying nuclear weapons, which means that k = 9

9 are destroyed, which means that n = 9

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

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

In which

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

P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021

P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

4 0
3 years ago
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