Software engineers are working on developing a fully automated fingerprint identification algorithm. Currently, expert examiners are required to identify the person who left the fingerprint. A study published in Psychological Science (August, 2011) tested the accuracy of experts and novices in identifying fingerprints. Participants were presented pairs of fingerprints and asked to judge whether the prints in each pair matched. The pairs were presented under three different conditions: prints from the same individual (match condition), non-matching but similar prints (similar distracter condition), and non-matching and very dissimilar prints (non-similar distracter condition). The percentages of correct decisions made by the two groups under each of the three conditions are listed in the table. Condition Fingerprint experts Novices
Match 92.12% 74.55%
Similar Distracter 99.32% 44.82%
Non-similar Distracter 100.00% 77.03%

Required:
a. Given a pair of matched prints, what is the probability that an expert will fail to identify the match?
b. Given a pair of matched prints, what is the probability that a novice will fail to identify the match?
c. Assume the study included 10 participants, 5 experts and 5 novices. Suppose that a pair of matched prints are presented to a randomly selected study participant and the participant fails to identify the match. Is the participant more likely to be an expert or a novice?

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

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step-by-step explanation:

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alright well the answers to your question is

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step-by-step explanation:

step-by-step explanation:

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