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Airida [17]
3 years ago
11

Consider the general form of the Reynolds transport theorem (RTT) given by dBsys dt = d dt ∫CV rhob dV + ∫CS rhobV› r·n › dA whe

re V › r is the velocity of the fluid relative to the control surface. Let Bsys be the mass m of a closed system of fluid particles. We know that for a system, dm/dt = 0 since no mass can enter or leave the system by definition. Use the given equation to derive the equation of conservation of mass for a control volume.

Engineering
1 answer:
musickatia [10]3 years ago
6 0

Answer:

Explanation:

note:

<u><em> solution is attached due to error in mathematical equation. please find the attachment</em></u>

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In order to write a good problem statement, you need to identify the following components of the problem.
forsale [732]

Answer:

True.

Explanation:

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An instance of the stable marriage problem has a unique stable matching if and only if the version of the Gale-Shapely algorithm
inysia [295]

Answer: True

Explanation:

The statement by Gale-Shapely algorithm is true where the male proposes and the version where female proposes both are correct meaning that altogether it is true. This is applicable when both yield the exact same match.

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3 years ago
Explain and show work:
belka [17]

Answer:

  5

Explanation:

The sum of the digits of the number is ...

  (4+1+3)+(4+6+5)+(7+8+9) = 8+15+24 = 47

The sum of those digits is 4+7=11, and those digits sum to 1+1 = 2.

That is, the value of the number mod 9 (or 3) is 2.

The ones digit is odd, so the value of the number mod 2 is 1.

This combination of modulo values tells you the mod 6 result is 5.

_____

<em>Additional comment</em>

We can look at the (mod2, mod3) values of the numbers 0 to 5:

  0 ⇒ (0, 0)

  1 ⇒ (1, 1)

  2 ⇒ (0, 2)

  3 ⇒ (1, 0)

  4 ⇒ (0, 1)

  5 ⇒ (1, 2) . . . . the mod {2, 3} results we have for the number of interest.

This process of adding up the digits repeatedly is referred to as "casting out 9s." The result of it is the modulo 9 value of the number (with 0 mapped to 9). Checking the mod 9 result of arithmetic operations is one quick way to spot certain kinds of errors. It can also be used as part of a divisibility test for 3 or 9.

3 0
3 years ago
Quinn’s relatives relayed a story about putting on a headset and seeing a digital world that they could walk around in and explo
Kryger [21]

Answer:

I know it is C)Virtual reality

Explanation:

Look at the clues

story about putting on a headset ( virtual reality head set!)

seeing a digital world (A virtual reality world)

they could walk around in (Fake walking you are basically jogging in place)

explore in order to see what ancient Benin looked like (Looking at a real place only digitally)

as if they were really there ( they think they are actually there)

The only reason I know all of this is because I have done virtual reality multiple times and I LOVED it SUPER fun ( I was doing archery) :) Hope this helps!

6 0
3 years ago
Read 2 more answers
A civil engineer is analyzing the compressive strength of concrete. The compressive strength is approximately normal distributed
hram777 [196]

Answer:

See explanation

Explanation:

Solution:-

- A study on compressive strength of a concrete was made. The distribution of compressive strength ( experimental testing ) was normally distributed with variance ( σ^2 ).

- A random sample of n = 12 specimens were taken and the mean compressive strength ( μ ) of 3500 psi was claimed.

- We are to test the claim made by the civil engineer regarding the mean compressive strength of the concrete. The data of compressive strength of each specimen from the sample is given below:

            3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237,

                          3286, 3210, 3265, 3273

- We will conduct the hypothesis whether the mean compressive strength of the concrete conforms to the claimed value.

      Null hypothesis: μ = 3500 psi

      Alternate hypothesis: μ ≠ 3500 psi

- The type of test performed on the sample data will depend on the application of Central Limit Theorem.

- The theorem states that the sample can be assumed to be normally distributed if drawn from a normally distributed population. ( We are given the population is normally distributed; hence, theorem applies )

- We will approximate the mean of the population ( μ ) with the sample mean ( x ), as per the implication specified by the theorem.

- The mean of the sample ( x ) is calculated as follows:

    x = \frac{Sum ( x_i )}{n} \\\\x = \frac{Sum ( 3273+ 3229+ 3256+ 3272+ 3201+ 3247+ 3267+ 3237+ 3286+ 3210+ 3265+3273 )}{12} \\\\x = \frac{39016}{12} \\\\x = 3251.3333

 

- Since, we are testing the average compressive strength of a concrete against a claimed value. Any value that deviates significantly from the claimed value is rejected. This corroborates the use of one sample two tailed test.

- The test value may be evaluated from either z or t distribution. The conditions for z-test are given below:

  • The population variance is known OR sample size ( n ≥ 30 )    

- The population variance is known; hence, we will use z-distribution to evaluate the testing value as follows:

              Z-test = \frac{x - u}{\sqrt{\frac{sigma^2}{n} } } \\\\Z-test = \frac{3251.333 - 3500}{\sqrt{\frac{1000^2}{12} } } \\\\Z-test = -27.24      

- The rejection region for the hypothesis is defined by the significance level ( α = 0.01 ). The Z-critical value ( limiting value for the rejection region ) is determined:

           Z-critical = Z_α/2 = Z_0.005

- Use the list of correlation of significance level ( α ) and critical values of Z to determine:

          Z-critical = Z_0.005 = ± 2.576

- Compare the Z-test value against the rejection region defined by the Z-critical value.

     Rejection region: Z > 2.576 or Z < -2.576

- The Z-test value lies in the rejection region:

            Z-test < Z-critical

           -27.24 < -2.576 .... Null hypothesis rejected

Conclusion: The claim made by the civil engineer has little or no statistical evidence as per the sample data available; hence, the average compressive strength is not 3500 psi.

- To construct a confidence interval for the mean compressive strength ( μ ) we need to determine the margin of error for the population.

- The margin of error (ME) is defined by the following formula:

              ME = Z^*. \frac{sigma}{\sqrt{n} }

Where,

- The ( Z* ) is the critical value for the defined confidence level ( CI ):

- The confidence interval and significance level are related and critical value Z* is as such:

   

            α = 1 - CI , Z* = Z_α/2

- The critical values for ( CI = 99% & 95% ) are evaluated:

           α = 1 - 0.99 = 0.01 , α = 1 - 0.95 = 0.05

           Z* = Z_0.005        ,   Z* = Z_0.025

           Z* = ± 2.58            ,   Z* = ± 1.96

- The formulation of Confidence interval is given by the following inequality:

                 [ x - ME  <    μ    <   x + ME ]

                 [ x - Z*√σ^2 / n   <    μ    <   x + Z*√σ^2 / n ]

- The CI of 95% yields:

   [ 3251.33 - 1.96*√(1000 / 12)   <    μ    <   3251.33 + 1.96*√(1000 / 12) ]

                [ 3251.333 - 17.89227 <    μ    <   3251.33 + 17.89227 ]

                              [ 3233.44  <    μ    <  3269.23  ]

- The CI of 99% yields:

   [ 3251.33 - 2.58*√(1000 / 12)   <    μ    <   3251.33 + 2.58*√(1000 / 12) ]

                [ 3251.333 - 23.552 <    μ    <   3251.33 + 23.552 ]

                              [ 3227.78  <    μ    <  3274.88  ]

                 

- We see that the width of the confidence interval increases as the confidence level ( CI ) increases. This is due to the increase in critical value ( Z* ) associated with the significance level ( α ) increases.    

7 0
3 years ago
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