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a_sh-v [17]
3 years ago
15

Certain items are purchased jointly. If each person pays 8 coins, the surplus is 3 coins, and if each person gives 7 coins, the

deficiency is 4 coins. Let x represent the number of people and y the number of coins needed to purchase the items. Find the number of people and the number of coins used to purchase the items.
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0

Answer:

7 people purchase the items and they used 53 coins

Step-by-step explanation:

Let x represent the number of people.

Let y be the number of coins needed to purchase the items.

If each person pays 8 coins, the surplus is 3 coins. This is illustrated below:

8x = y + 3 (1)

if each person gives 7 coins, the deficiency is 4 coins. This is illustrated below:

7x = y — 4 (2)

Solving by elimination method: subtract equation(2) from equation (1). This is illustrated below:

8x = y + 3 (1)

— (7x = y — 4) (2)

x = 7.

Next, Substituting the value of x into any of the equation to obtain y. In this case I will be using equation 1 as illustrated below:

8x = y + 3 (1)

8(7) = y + 3

56 = y + 3

Collect like terms

y = 56 — 3

y = 53

Therefore, 7 people purchase the items and they used 53 coins

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1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n
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Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 70% of fatalities involve an intoxicated driver, hence p = 0.7.
  • A sample of 15 fatalities is taken, hence n = 15.

The probability is:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

Hence

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

P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061

P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186

P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700

P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916

P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305

P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047

Then:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0
3 years ago
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