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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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A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten
kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

\mu = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

\omega = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

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3 years ago
Which is the following is made up mostly of water
tatyana61 [14]

soil i think is this from apex?

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3 years ago
A kite 100 ft above the ground moves horizontally at a speed of 12 ft/s. at what rate is the angle (in radians) between the stri
Viefleur [7K]
<span>Answer: So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z. So we have x2 + 1002 = z2. Now take its derivative in terms of time to get 2x(dx/dt) = 2z(dz/dt) So at your specific moment z = 200, x = 100âš3 and dx/dt = +8 substituting, that makes dz/dt = 800âš3 / 200 or 4âš3. Part 2 sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get cos a (da./dt) = -100/ z2 (dz/dt) So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or Ď€/6 radians. Substitute to get cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3) âš3 / 2 (da/dt) = -âš3 / 100 da/dt = -1/50 radians</span>
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3 years ago
A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from
o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}

There is NO initial vertical velocity, so:

\large\boxed{\Delta d= \frac{1}{2}at^2}}

Rearrange to solve for time:

2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}

6 0
2 years ago
The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called lase
emmainna [20.7K]

The answer is photocoagulation.

The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called laser <u>photocoagulation.</u>

<u></u>

What is photocoagulation?

A minimally invasive method used to treat numerous retinal illnesses is photocoagulation of the retina, also known as retinal laser photocoagulation. The retina may expand due to aberrant leaky blood vessels developing across it in a number of disorders. Laser photocoagulation uses thermal energy above 65 °C to burn the retinal tissue by creating thermal burns. This can prevent the retina from being damaged by the bleeding blood vessels. In addition to causing fibrosis, laser photocoagulation can also seal retinal tears. Laser photocoagulation is typically unable to recover already lost vision in cases of retinal disease, but it can slow the progression of the condition, lower the chance of further vision loss, and preserve residual vision. The likelihood of problems following the operation is quite minimal.

To learn more about photocoagulation click on the link below:

brainly.com/question/16016898

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6 0
1 year ago
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