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Dvinal [7]
3 years ago
8

You have just timed a person doing a hair cut for the first time. It took 50 minutes. What unit improvement factor learning curv

e would you use if the person took 35 minutes on the second hair cut?
A. 35 percent
B. 50 percent
C. 70 percent
D. 75 percent
E. 80 percent
Mathematics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer: C. 70 percent

Step-by-step explanation:

Given, Time for the first unit = 50 minutes

Time for the second unit = 35 minutes

The unit improvement factor learning curve = (The time for the second unit) ÷ (time for the first unit) x 100.

So, The unit improvement factor learning curve  = 35÷ 50 × 100 = 70 percent.

Hence, the correct option is "C. 70 percent".

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Newton's law of cooling is:
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Answer:

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

\frac{du}{dt}= -k (u-T)

We can reorder the expression like this:

\frac{du}{u-T} = -k dt

We can use the substitution w = u-T and dw =du so then we have:

\frac{dw}{w} =-k dt

IF we integrate both sides we got:

ln |w| = -kt +C

If we apply exponential in both sides we got:

w = e^{-kt} *e^c

And if we replace w = u-T we got:

u(t)= T + C_1 e^{-kt}

We can also express the solution in the following terms:

u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}

For this case we know that k =-0.15 hr since w ehave a cooloing, T_{i}= 70 F, T_{amb}=11F, we have this model:

u(t) = (70-11) e^{-0.15t} +11

And if we want that the temperature would be 32F we can solve for t like this:

32 = 59 e^{-0.15 t} +11

21=59 e^{-0.15 t}

\frac{21}{59} = e^{-0.15 t}

If we apply natural logs on both sides we got:

ln (\frac{21}{59}) =-0.15 t

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So it would takes approximately 6.9 hours to reach 32 F.

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4 years ago
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