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3 years ago

C= 7/6 (k-17) Solve for k.

Mathematics

1 answer:

Fed [463]3 years ago

7 0

Answer:

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Step-by-step explanation:

c= 7/6k- 17(7/6)

6/7c = k - 17

k = 6/7c+17

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Newton’s law of cooling states that dx/dt= −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k

Vadim26 [7]

Answer:

(a) The solution to the differential equation is x = A_0Coswt + Ce^(-kx)

(b) The initial condition t > 0 will not make much of a difference.

Step-by-step explanation:

Given the differential equation

dx/dt= −k(x − A); t > 0, A = A_0Coswt

(a) To solve the differential equation, first separate the variables.

dx/(x - A) = -kdt

Integrating both sides, we have

ln(x - A) = -kt + c

x - A = Ce^(-kt) (Where C = ce^(-kt))

x = A + Ce^(-kx)

Now, we put A = A_0Coswt

x = A_0Coswt + Ce^(-kx) (Where C is constant.)

And we have the solution.

(b) Since temperature t ≠ 0, the initial condition t > 0 will not make much of a difference because, Cos(wt) = Cos(-wt).

It is not any different from when t < 0.

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Answer:

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A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

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Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

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Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

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So, standard deviation of 1 car is given as:

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Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

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Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

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C= 7/6 (k-17) Solve for k. ​

C= 7/6 (k-17) Solve for k.