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Liula
2 months ago
11

Mr. Itol's assistant takes twice as long to complete a computer task as Mr. Itol. If it takes both experts 6 hours to complete t

he task, how long will it take each of them to do the job alone?
Mathematics
1 answer:
Inessa [12.5K]2 months ago
4 0
To tackle this problem, I will formulate an equation. Let's designate x as the number of hours Mr. Itol requires to finish a computer task, while since his assistant needs twice as long, we'll set 2x as the hours the assistant needs. We also know that together they complete the task in 6 hours, allowing us to establish the following equation:

\frac{1}{2x} + \frac{1}{x} = \frac{1}{6}

Now, to solve this, we need to identify the least common denominator, allowing us to eliminate the denominators. The LCD is 6x, which modifies our equation to:

\frac{3}{6x} + \frac{6}{6x} = \frac{x}{6x}

Next, we can eliminate the denominators, resulting in:

3+6=x

Thus, x = 9.

This tells us that Mr. Itol would take 9 hours to work on the task alone. To calculate the duration needed for his assistant to complete it independently, we substitute the x-value into 2x:

2(9)=18

Consequently, the final determination is that Mr. Itol would need 9 hours to finish the task alone, while his assistant would require 18 hours (twice as long) to do so independently.

Hopefully, this is clear!:)










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The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable
Zina [12379]

Answer:

(A) 0.15625

(B) 0.1875

(C) Cannot be determined

Step-by-step explanation:

The time it takes for a student to finish a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let's denote X as the duration needed for the student to complete the statistics quiz

Thus, X ~ U(32, 64)

The probability density function (PDF) for a uniform distribution is expressed as;

f(X) = \frac{1}{b-a},  a < X < b      where a = 32 and b = 64

The cumulative distribution function (CDF) is given by P(X <= x) = \frac{x-a}{b-a}

(A) The probability of a student taking longer than 59 minutes to complete the quiz = P(X > 59)

   P(X > 59) = 1 - P(X <= 59) = 1 - \frac{x-a}{b-a} = 1 - \frac{59-32}{64-32} = 1-\frac{27}{32} = 0.15625

(B) The probability that a student completes the quiz between 37 and 43 minutes = P(37 <= X <= 43)  = P(X <= 43) - P(X < 37)

    P(X <= 43) = \frac{43-32}{64-32} = \frac{11}{32} = 0.34375

    P(X < 37) = \frac{37-32}{64-32} = \frac{5}{32} = 0.15625

    P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

(C) The probability that a student takes exactly 44.74 minutes to complete the quiz

     = P(X = 44.74)

This probability cannot be calculated as it is a continuous distribution, which doesn't provide probabilities for specific points.

3 0
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A recent review of a compact disc distributor’s product line is summarized as follows: Selling Price ($) Number of Titles Midpoi
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I'm looking for the same answer. If you have found any solutions, please share them with me!!

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when joe bowls he can get a strike 60% of the time, what is the probability jow will get at least 3 strikes out of 4 tries?
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34.56%. This is a binomial probability that can efficiently be calculated using the following formula: Here, n signifies the total number of trials (in this case, 4), x denotes the number of "successes" (which is 3), p is the success probability (60% or 0.6), and q indicates the failure rate (1 - p, thus 0.4). Plugging these values into the formula yields the solution: in percentage form, the probability is found to be 34.56%.
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Bob Nale is the owner of Nale's Texaco GasTown. Bob would like to estimate the mean number of litres (L) of gasoline sold to his
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a. The point estimate for the population mean is b. The confidence interval at 80% is c. This means there is an 80% probability that the true mean of the population lies within the given confidence interval.
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