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bija089 [108]
4 years ago
8

Greg spent $4 on 5 packs of sport cards.

Mathematics
1 answer:
tino4ka555 [31]4 years ago
5 0

Answer:

(a) Money spend on each pack will be $0.8

(b) Money send on each pack will be $0.2

Step-by-step explanation:

We have given that Greg spent $4 on 5 packs of spot card

(a) We have to find how much Greg spend on each pack

Cost of each pack will be =\frac{4}{5}=$0.8

(b) In this case it is said that Greg spent half of money

So money spent by Greg =\frac{4}{2}=$2

And he bought twice of pack than earlier

So number of packs he bought =5\times 2=10

So money spend on each pack will be equal to =\frac{2}{10}=$0.2

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On a particular day, 112 of 280 passengers on a particular DTW-LAX flight used the e-ticket check-in kiosk to obtain boarding pa
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Answer:

0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

Step-by-step explanation:

Hypergeometric distribution:

The probability of x sucesses 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 sucesses.

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:

Population of 280, which means that N = 280

112 use e-ticket check-in kiosk, which means that k = 112

Sample of eight passengers means that n = 8

Find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

This is P(X = 4). So

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

P(X = 4) = h(4,280,8,112) = \frac{C_{112,4}*C_{168,4}}{C_{280,4}} = 0.2348

0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

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