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frosja888 [35]
1 year ago
6

Which of the following is a correct way to measure productivity? a. Divide the number of hours worked by the quantity of output.

b. Divide the quantity of output by the number of hours worked. c. Determine how much output is produced in a given time. d. Determine how much time it takes to produce a unit of output.
Mathematics
1 answer:
11111nata11111 [884]1 year ago
6 0

Answer:

b. Divide the quantity of output by the number of hours worked.

Step-by-step explanation:

<em>Since the ratio of the number of output to the number of hours worked shows the productivity. </em>

Thus, option (b) is correct.

Productivity is used to converting inputs into useful output. It measures the efficiency of a person, system, machine, factory, etc.

For Example: The employee who works less hours and assembled more radios has more productivity, that employee knows how to utilize time.

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A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556
MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then,  \overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560

Now, \sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76

Then, \sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76

Hence, the  standard deviation of his measurements = 2.76 g

6 0
2 years ago
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23 +30 (15 feet each way) = 53

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Hello!

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