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Anit [1.1K]
3 years ago
10

Calculate the first and second velocities of the car with three washers attached to the pulley, using the formulas v1 = 0. 25 m

/ t1, and v2 = 0. 25 m / (t2 – t1) where t1 and t2 are the average times the car took to reach the 0. 25 and the 0. 50 meter marks. Record these velocities, to two decimal places, in Table E. What is the first velocity of the car with three washers at the 0. 25 meter mark? m/s What is the second velocity of the car with three washers at the 0. 50 meter mark? m/s.
Mathematics
1 answer:
Travka [436]3 years ago
3 0

The first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

Given data:

The value of the first distance is, d1 = 0.25 m.

The value of the second distance is, d2 = 0.50 m.

<h3>What is velocity?</h3>

The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,

v = d/t

Here, d is the distance covered.

And t is the average time taken to cover the distance.

Then the first velocity of the car at 0.25 m is,

v1 = d1/t1

v1 = 0.25 / t1

Here, t1 is the average time for the first distance.

And the second velocity of the car with four washers at the 0. 50 m mark is,

v2 = d2/t2

v2 = 0.50 /t2

here, t2 is the average time for the second distance.

Thus, we can conclude that the first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

To know more about Velocity click the link given below.

brainly.com/question/25796948?

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Given : The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0, 0.3, 0.4, and 0.6, respectively.

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Let the event A be the no printer problem i.e. P(A)=0.8

Let the event B be the misaligned paper i.e. P(B)=0.02

Let the event C be the high ink viscosity i.e. P(C)=0.08

Let the event D be the printer-head debris i.e. P(D)=0.1

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P(E)=P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)+P(D)P(E|D)

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a. Determine the probability of high ink viscosity given poor print quality.

P(C|E)=\frac{P(E|C)P(C)}{P(E)}

P(C|E)=\frac{0.4\times 0.08}{0.098}

P(C|E)=\frac{0.032}{0.098}

P(C|E)=0.3265

b. Given poor print quality, what problem is most likely?

Probability of no printer problem given poor quality is

P(A|E)=\frac{P(E|A)P(A)}{P(E)}

P(A|E)=\frac{0\times 0.8}{0.098}

P(A|E)=\frac{0}{0.098}

P(A|E)=0

Probability of misaligned paper given poor quality is

P(B|E)=\frac{P(E|B)P(B)}{P(E)}

P(B|E)=\frac{0.3\times 0.02}{0.098}

P(B|E)=\frac{0.006}{0.098}

P(B|E)=0.0612

Probability of printer-head debris given poor quality is

P(D|E)=\frac{P(E|D)P(D)}{P(E)}

P(D|E)=\frac{0.6\times 0.1}{0.098}

P(D|E)=\frac{0.06}{0.098}

P(D|E)=0.6122

From the above conditional probabilities,

The printer-head debris problem is most likely given that print quality is poor.

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