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Luba_88 [7]
3 years ago
15

Suppose you're studying levels of job stress and measures of job efficiency in a nursing home. Which one of the following statem

ents best represents a data comparison that demonstrates a positive correlation
A. As stress levels increase, job efficiency increases.
B. As job efficiency decreases, stress levels increase.
C. As job efficiency increases, stress levels decline.
D. As stress levels increase, job efficiency declines
Physics
2 answers:
zloy xaker [14]3 years ago
7 0

Answer: Hello mate!

You are studying the relationship between levels of job stress and measures of job efficiency.

now, we want to describe a positive correlation.

A correlation between two quantities, A and B, is A = kB

where k is a constant.

If the correlation is positive, means that when A increases also increase B, and then k is a positive number.

This means that, in our case,  when job stress increases, job efficiency also increases.

Then the right answer is A "As stress levels increase, job efficiency increases."  

coldgirl [10]3 years ago
6 0
B. As job efficiency increases, stress levels decline (THIS IS WRONG)I think its between A. As job efficiency decreases, stress levels increaseD. As stress levels increase, job efficiency increases
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Read 2 more answers
A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi
sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

7 0
3 years ago
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