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Leona [35]
3 years ago
9

Tapiwa raked 5% more leaves than Adam raked. Together, they raked 697 liters of leaves. How many liters of leaves did Adam rake?

Mathematics
1 answer:
EleoNora [17]3 years ago
7 0
Assume that the number of leaves raked by Adam is x.
Tapiwa raked 5% more leaves than Adam, this means that:
Leaves raked by Tapiwa = x + 0.05x = 1.05x liters
We are given that the total number of raked leaves is 697 liters.
This means that:
Total raked = raked by Adam + raked by Tapiwa
697 = x + 1.05x
697 = 2.05x
x = 697 / 2.05 
x = 340
Based on the above calculations:
Adam raked 340 liters of leaves
Tapiwa raked 1.05(340) = 357 liters of leaves
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Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which 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)!}

And p is the probability of X happening.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is p = \frac{1}{5} = 0.2

10 questions.

This means that n = 10

A) What is the probability that he will answer all questions correctly?

This is P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so P(X = 0)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

P(X \geq 1) = 1 - P(X = 0)

Since P(X = 0) = 0.1074, from item b.

P(X \geq 1) = 1 - 0.1074 = 0.8926

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264

P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055

P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008

P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001

P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0

P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0

So

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

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