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Digiron [165]
2 years ago
6

Newton’s law of cooling states that for a cooling substance with initial temperature T0 , the temperature T(t) after t minutes c

an be modeled by the equation T(t)=Ts+(T0−Ts)e−kt , where Ts is the surrounding temperature and k is the substance’s cooling rate. A liquid substance is heated to 80°C . Upon being removed from the heat, it cools to 60°C in 25 min. What is the substance’s cooling rate when the surrounding air temperature is 50°C ? Round the answer to four decimal places.
Mathematics
1 answer:
daser333 [38]2 years ago
4 0
T(t) = Ts + (T₀ - Ts)exp(-kt), substituting the values:
60 = 50 + (80 - 50)exp(-25k)
10/30 = exp(-25k)
k = 0.0439 °C/min
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Answer:​

The correct answer is No.

Choosing between the critical value method or the P-value method does not affect one's conclusion because both methods look at the probability of the test​ statistic's and its level of significance .

Given the methodology utilized by both methods, they usually arrive at the same conclusion.

Cheers!

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Find a single expression that represents the area of the outer ring of the circle if the area of the whole circle is represented
sp2606 [1]

Answer:  The answer is A_o=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


Step-by-step explanation: Given that the area of the whole circle is represented by the expression

A_c=25x^2-12x-9.

We are to find the area of the outer ring of the circle, i.e., to find the circumference of the circle.

Now, if 'r' represents the radius of the circle, then we have

A_c=\pi r^2\\\\\Rightarrow \dfrac{22}{7}r^2=25x^2-12x-9\\\\ \Rightarrow r^2=\dfrac{7}{22}(25x^2-12x-9)\\\\\Rightarrow r=\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.

Thus, the area of the outer ring is

A_o=2\pi r=2\pi\sqrt{\dfrac{7}{22}(25x^2-12x-9)}.


5 0
2 years ago
The population can be modeled by P(t) = 82.5 − 67.5cos⎡ ⎣(π/6)t ⎤ ⎦, where t is time in months (t = 0 represents January 1) and
Fed [463]

Answer:

The intervals in which the population is less than 20,000 include

(0 ≤ t < 0.74) and (11.26 < t ≤ 12)

Step-by-step explanation:

P(t) = 82.5 - 67.5 cos [(π/6)t]

where

P = population in thousands.

t = time in months.

During a year, in what intervals is the population less than 20,000?

That is, during (0 ≤ t ≤ 12), when is (P < 20)

82.5 - 67.5 cos [(π/6)t] < 20

- 67.5 cos [(π/6)t] < 20 - 82.5

-67.5 cos [(π/6)t] < -62.5

Dividing both sides by (-67.5) changes the inequality sign

cos [(π/6)t] > (62.5/67.5)

Cos [(π/6)t] > 0.9259

Note: cos 22.2° = 0.9259 = cos (0.1233π) or cos 337.8° = cos (1.8767π) = 0.9259

If cos (0.1233π) = 0.9259

Cos [(π/6)t] > cos (0.1233π)

Since (cos θ) is a decreasing function, as θ increases in the first quadrant

(π/6)t < 0.1233π

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t < 6×0.1233

t < 0.74 months

If cos (1.8767π) = 0.9259

Cos [(π/6)t] > cos (1.8767π)

cos θ is an increasing function, as θ increases in the 4th quadrant,

[(π/6)t] > 1.8767π (as long as (π/6)t < 2π, that is t ≤ 12)

(t/6) > 1.8767

t > 6 × 1.8767

t > 11.26

Second interval is 11.26 < t ≤ 12.

Hope this Helps!!!

3 0
2 years ago
Question 2 (1 point)
hichkok12 [17]

Answer:

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