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Kazeer [188]
1 year ago
5

Giving a test to a group of students, the grades and gender are summarized belowGrades vs. Gender ABCMale10316Female465If one st

udent was chosen at random, find the probability that the student got a B.
Physics
1 answer:
strojnjashka [21]1 year ago
5 0

Answer:

20.45%

Explanation:

The probability that the student got a B is

\frac{total\#of\text{ students that got B}}{\text{total students }}\times100\%

Now, how many students are there in total?

The answer is

10+3+16+4+6+5=44\; \text{students}

How many students got a B?

The answer is

3+6=9\; \text{students}

therefore, the probability that the student has got a B is

\frac{9\text{ students }}{44\text{ students }}\times100\%=20.45\%

Hence, the probability that a student has got a B is 20.45%

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Which of the following planets rotates in the opposite direction than the other planets? Venus Neptune Saturn Uranus
pogonyaev
Hello,

The answer is option A "Venus".

Reason:

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3 years ago
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A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done o
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Answer:

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5 0
2 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
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