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schepotkina [342]
3 years ago
13

Three judges measure the lap times of a runner and each measurement is different from the others by a few hundredths of a second

. The most likely reason for this difference is
Mathematics
1 answer:
ohaa [14]3 years ago
6 0

Answer:

Systematic error

Step-by-step explanation:

Assuming that none of the judges are biased, the most likely reason for this difference is the occurrence of systematic errors.

Systematic errors are errors introduced by inaccuracy in the experimental design, be it in the observation or measurement process.

In this case, the reaction time from observing the finish and stopping the clock for each judge might be different, which configures a systematic error.

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It increases

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(5/x) + 5

As x decreases, 5/x increases.  So the expression increases.

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Len [333]

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48%

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To convert any fraction to a percentage, we need to multiply it by 100.

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An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same
elena-s [515]

Answer:

Approximately 0.15 (360 / 2401.) (Assume that the choices of the 5 passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all 7 floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these 5 passenger could choose from any one of the 7 floors. There would be a total of 7 \times 7 \times 7 \times 7 \times 7 = 7^{5} unique ways for these 5\! passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the 7 floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only (7 - 1) = 6 floors.

Likewise, the third passenger would have to choose from only (7 - 2) = 5 floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only (7 \times 6 \times 5 \times 4 \times 3) unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the 7 floors. Each of these 7^{5} combinations would be equally likely.

Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:

\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}.

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C

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Veseljchak [2.6K]

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-4 and -2

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(X+4)(x+2)

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