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Irina18 [472]
3 years ago
9

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

arton, T, as a function of time follows Newton's cooling law:
dT / dt = k (T - a),
where a is the temperature of the surroundings and k is a constant. We assume the temperature
the surroundings are a = 20 (given in ◦C), and that T (0) = 4 (given in ◦C).
a) Show that T (t) = 20 - 16e ^ kt (given in ◦C).
b) After 5 minutes the temperature of the carton is 8◦C. What is the temperature after 15 minutes?
Mathematics
1 answer:
Kisachek [45]3 years ago
6 0
We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

\frac{dT}{dt}=k(T-a)

We now separate the time and the temperature variables as follows,

\frac{dT}{T-a}=kdt

Integrating both sides of the differential equation, we obtain;

ln(T-a)=kt +c

This natural logarithmic equation can be rewritten as;

T-a=e^{kt +c}

Applying the laws of exponents, we obtain,

T-a=e^{kt}\times e^{c}

T-a=e^{c}e^{kt}

We were given the initial conditions,

T(0)=4

Let us apply this condition to obtain;

4-20=e^{c}e^{k(0)}

-16=e^{c}

Now our equation, becomes

T-a=-16e^{kt}

or

T=a-16e^{kt}

When we substitute a=20,
we obtain,

T=20-16e^{kt}

b) We were also given that,

T(5)=8

Let us apply this condition again to find k.

8=20-16e^{5k}

This implied

-12=-16e^{5k}

\frac{-12}{-16}=e^{5k}

\frac{3}{4}=e^{5k}

We take logarithm to base e of both sides,

ln(\frac{3}{4})=5k

This implies that,

\frac{ln(\frac{3}{4})}{5}=k


k=-0.2877

After 15 minutes, the temperature will be,


T=20-16e^{-0.2877\times 15}

T=20-0.21376


T=19.786

After 15 minutes, the temperature is approximately 20°C


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Answer:

The bottom for an A is 75. Round to the nearest whole number as needed.

Step-by-step explanation:

1) Previous concepts

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The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the scores on this case, and for this case we know the distribution for X is given by:

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And we want the top 5.5% of the scores, so we need a value a such that:

P(X>a)=0.055 or P(X

We need on the right tail of the distribution a value a that gives to us 94.5% of the area below and 5.5% of the area above. Both conditions are equivalent.

Let's use the condition P(X, the best way to solve this problem is using the z score with the following formula:

z=\frac{x-\mu}{\sigma}

So we need a value from the normal standard distribution that accumulates 0.945 of the area on the left and 0.055 on the right. This value on this case is 1.598 and we can founded with the following code in excel:

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If we apply the z score formula to our case we have this:

P(X

So then based on the equalities we have this:

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Answer:

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Answer:

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Answer:

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