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4 years ago

Please help!!!!!! please please please!!!

Mathematics

1 answer:

SVETLANKA909090 [29]4 years ago

4 0

Answer:

290

Step-by-step explanation:

These are rectangular prisms. The volume formula for a rectangular prism is

l x w x h.

I'll call the bottom cuboid 'Figure A'.

Figure A has a length of 8, a width of 5, and a height of 5. The volume formula is: 8 x 5 x 5. 40 x 5 =200

The volume of Figure A is 200.

I'll call the top cuboid Figure B.

Figure B has a length of 3, a width of 5, and a height of 6. That turns into

3 x 5 x 6.

The volume of Figure B is 90.

The volume of a composite figure is the sum of the figure's volume.

200+90=290

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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

Sonja [21]

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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3 years ago