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Answer:

Time constant = 15.34 seconds

The thermometer indicates an error margin of 0.838°

Explanation:

Given

t = 1 minute = 60 seconds

c(t) = 98% = 0.98

According to the details provided, the thermometer functions as a first-order system.

The transfer function of such a system is expressed as;

C(s)/R(s) = 1/(sT + 1).

To determine the time constant, the step response must be evaluated.

This is defined as

r(t) = u(t) --- Applying Laplace Transformation

R(s) = 1/s

Replacing 1/s back into C(s)/R(s) = 1/(sT + 1).

What we have

C(s)/1/s = 1/(sT + 1)

C(s) = 1/(sT + 1) * 1/s

C(s) = 1/s - 1/(s + 1/T) --- Taking Inverse Laplace Transformation

L^-1(C(s)) = L^-1(1/s - 1/(s + 1/T))

Given that e^-t <–> 1/(s + 1) --- {L}

1 <–> 1/s {L}

Thus, the unit response c(t) = 1 - e^-(t/T)

Substituting 0.98 for c(t) and 60 for t

0.98 = 1 - e^-(60/T)

0.98 - 1 = - e^-(60/T)

-0.02 = - e^-(60/T)

e^-(60/T) = 0.02

ln(e^-(60/T)) = ln(0.02)

-60/T = -3.912

T = -60/-3.912

T = 15.34 seconds

It follows that the time constant = 15.34 seconds

The error signal is characterized by

E(s) = R(s) - C(s)

Where the temperature shifts at 10°/min; which equals 10°/60 s = 1/6

Thus,

E(s) = R(s) - 1/6 C(s)

Calculating C(s)

C(s) = 1/s - 1/(s + 1/T)

C(s) = 1/s - 1/(s + 1/15.34)

Note that R(s) = 1/s

Thus, E(s) turns into

E(s) = 1/s - 1/6(1/s - 1/(s + 1/15.34))

E(s) = 1/s - 1/6(1/s - 1/(s + 0.0652)

E(s) = 1/s - 1/6s + 1/(6(s+0.0652))

E(s) = 5/6s + 1/(6(s+0.0652))

E(s) = 0.833/s + 1/(6(s+0.0652)) ---- Taking Inverse Laplace Transformation

e(t) = 1/6e^-0.652t + 0.833

In a first-order system, a steady state condition is reached when the time is four times the time constant.

Thus,

Time = 4 * 15.34

Time = 61.36 seconds

Consequently, e(t) becomes

e(t) = 1/6e^-0.652t + 0.833

e(t) = 1/(6e^-0.652(61.36)) + 0.833

e(t) = 0.83821342824942664566211

e(t) = 0.838 --- Rounded off

Thus, the thermometer reveals an error of 0.838°

Answer:

The response is presented below

Explanation:

Central Limit Theorem: The CLT is a statistical principle asserting that if a sample size is sufficiently large and shows minimal variation from the population, the average of all samples will approximate the mean of that population. Furthermore, all distributions can be modeled as nearly normal.

This principle is illustrated by the central limit theorem. Specifically, if we gather independent random samples of size n from any kind of population, a large n will yield a sample distribution that aligns with a normal distribution pattern.

mean of the sample means

                    .............1

while the standard deviation of the sample means is

                      .........2

and This theory has findings applicable elsewhere in statistical studies. Although it may initially seem theoretical and lacking real-world utility, the Central Limit Theorem is crucial for applying statistics effectively in practice.

Answer:

Han would share new insights along with reasoning for his agreement or disagreement with Emily. This could prompt other participants to engage with Han's thoughts.

Hope this helps!~ ♡

Response:

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Clarification:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

Assuming depth of flow as "d"

Q = 1/n (R)^2/3 (slope)^1/2 A   ( R represents the hydraulic Radius)

By inserting the provided data into the above equation:

400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d

R = A/P

In this case, A stands for the flow area and P indicates the wetted perimeter

400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d

d = 1.42 feet

Depth of stilling channel before the jump = d1 = 8 feet

Depth of stilling channel after the jump = d2 = 1.42 feet

Length of stilling basin = 5(d2 - d1)

= 5( 8 - 1.42)

Length of stilling basin = 32.9 feet

Now calculating the height of end sill:

Jump height = (8 - 1.42)

Height of end sill = 6.58 feet