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TiliK225 [7]
3 years ago
14

Fred has 30 minutes to do a three-problem quiz. He spent 10 7/10 minutes on question a and 5 2/5 minutes on question b. How much

time did he have left for question c?
Mathematics
1 answer:
tatiyna3 years ago
3 0

Answer:Fred has 13 9/10 minutes left for question C

Step-by-step explanation:

The total time that Fred has to do the three-problem quiz is 30 minutes.

He spent 10 7/10 minutes on question A. Converting 10 7/10 minutes to improper fraction, it becomes 107/10 minutes.

He spent 5 2/5 minutes on question B. Converting 5 2/5 minutes to improper fraction, it becomes 27/5 minutes.

Total time that Fred spent on question A and question B is

107/10 + 27/5 = (107 + 54)/10

= 161/10 minutes.

The amount of time that he has left for question C would be

30 - 161/10 = (300 - 161)/10 = 139/10

= 13 9/10 minutes

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Answer:

<em>Part a) Probability that a moose in that age group is killed by a wolf</em>

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1) Start by arranging the table to interpret the information:

<u>Age of Moose in years          Number killed by woolves</u>

Calf (0.5)                                                 107

1 - 5                                                           47

6 - 10                                                         79

11 - 15                                                        60

16 - 20                                                        3

You can  now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.

2) <u>First question: </u>a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.

i) Formula:

Probability = number of positive outcomes / total number of events.

ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.

iii) P  (0.5 year) = 107 / 296 = 0.361

iv) P (1 - 5) = 47 / 296 = 0.159

v) P (6 - 10) = 79 / 296 = 0.267

vi) P (11 - 15) = 60 / 296 = 0.203

vii) P (16 - 20) = 3 / 296 = 0.010

3) <u>Second question:</u> b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.

i) Class       midpoint

   0.5               0.5

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   6 - 10         (6 + 10) / 2 = 8

   11 - 15         (11 + 15) / 2 = 13

   16 - 20       (16 + 20) = 18

ii) Expected age of a moose killed by a wolf = mean of the distribution = μ

μ = Sum of the products of each probability times its age (mid point)

μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years

μ = 5.61 years ← answer

iii) Stantard deviation of the ages = σ

σ = square root of the variace

variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.

s =  (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8  - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65

σ = √ (24.65) = 4.97 years ← answer

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