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vredina [299]
2 years ago
6

the dealer's coast of a car is 85% of the listed price. The dealer would accept any offer that is at least 500 dollar over the d

ealer's cost. design an algorithm that prmpts the user to input the list price of the car and print the least amount that the dealerr would accept for the car
Engineering
1 answer:
schepotkina [342]2 years ago
7 0

The dealer would accept any offer that is at least 500 dollars over the dealer's cost The algorithms for your query are 1) Declare list Price as a double READ list Price• Declare and initialize most-0.85*list Price+500.0.

<h3>What is an easy definition of a set of rules?</h3>

A set of rules is a method used for fixing a hassle or acting a computation. Algorithms act as a specific listing of commands that behavior certain moves grade by grade in both hardware- or software-primarily based totally routines. Algorithms are broadly used during all regions of IT.

  1. Function calculateAvg(Scores) Declare and initialize SUM as double identical to 0 • For i=1 to length(Scores) SUM=SUM+Scores.get(i).
  2. EndLoop- Return SUM/length(Scores)
  3. EndFunction- Function printBelowAvg(Names, Scores)
  4. Define and claim AVG=calculateAvg(Scores), For i=1 to length(Scores)
  5. If Scores.get(i)MAX then MAX=Scores.get(i)
  6. Endif,  EndLoop.
  7. RETURN MAX-EndFunction- Function blended Perform(Names, Scores).
  8. PRINT calculateAvg(Scores), print BelowAvg(Names, Scores)
  9. PRINT highestScore(Scores), EndFunction
  10. Call the characteristics as combinedPerform(Names, Scores)
  11. Kindly revert for any queries.

Read more about the algorithm :

brainly.com/question/24953880

#SPJ1

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The reaction at support B

Rb= 235440N

The reaction at support C

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Explanation : check attachment

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What is the built-in pollution control system in an incinerator called
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Which statement about lean manufacturing is true when you compare it to mass production?
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Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

6 0
4 years ago
9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space &amp; n(s)
zzz [600]

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

7 0
3 years ago
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