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Setler [38]
4 years ago
14

Select the correct answer from the drop-down menu. A box contains shirts in two different colors and two different sizes. The nu

mbers of shirts of each color and size are given in the table. Shirt Color Size Large Medium Total Red 42 48 90 Blue 35 40 75 Total 77 88 165 From the data given in the table, we can infer that .
Physics
2 answers:
kogti [31]4 years ago
8 0

Answer:

Plato Users: its P(blue shirt l large shirt) = P(blue shirt)

Explanation:

yKpoI14uk [10]4 years ago
7 0

Answer:

P( red | large shirt ) ≠ P( large shirt)

P( Blue | large shirt ) = P( blue shirt)

P( shirt is medium and blue ) ≠ P( medium  shirt)

P( large shirt | red shirt ) ≠ P( red shirt)

Explanation:

P( red | large shirt ) = 42/77 = 0.5454 : P( large shirt)= 77/165 = 0.467

P( red | large shirt ) ≠ P( large shirt)

P( Blue | large shirt ) = 35/77 = 0.4545 : P( blue shirt)= 75/165 = 0.4545

   P( Blue | large shirt ) = P( blue shirt)

P( Shirt is medium and blue ) = 40/77 = 0.2424   : P( medium shirt)= 88/165 = 0.5333

P( shirt is medium and blue ) ≠ P( medium  shirt)

P( large shirt | red shirt) = 42/90 = 0.4667 : P( red shirt)= 90/165 = 0.5454

P( large shirt | red shirt ) ≠ P( red shirt)

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Answer the point I wish you would have the greatest potential energy is when you are coming down the swing and getting ready to go up the greatest kinetic energy is whenever you’re falling back down from the height of how far you went up
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3 years ago
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A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
astraxan [27]

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

5 0
4 years ago
A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a spee
Elodia [21]

Answer:

a.2.5x 10^3 m/s

b.mr=48kg/s

Explanation:

A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a speed of 5.0 ✕ 103 m/s relative to the rocket. The mass of the rocket at this time is 6.0 ✕ 104 kg, and its acceleration is 4.0 m/s2. What is the velocity of the exhaust relative to the solar system? (B) At what rate was the exhaust ejected during the firing?

velocity of the exhaust relative to the solar system

velocity of the rocket -velocity of the exhaust relative to the rocket.

7.5 ✕ 103 m/s-5.0 ✕ 103 m/s

2.5x 10^3 m/s

. b  we will look for the thrust of the rocket

T=ma

T=6.0 ✕ 104 kg*4.0 m/s2

T=2.4*10^5N

f=mass rate *velocity of the exhaust

T=2.4*10^5N=mr*5.0 ✕ 10^3 m/s

mr=2.4*10^5N/5.0 ✕ 10^3

mr=48kg/s

5 0
3 years ago
Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of th
Mila [183]

Answer:

a. Picture attached

b. V, X

c. t=0.12 s

d. It makes sense cosidering that a normal heart beat rate is between 60 and 100 beats per minute (bpm).

Explanation:

First we have to identify the unknown of this problem:

a: acceleration\\t: time

Considering the characteristics of the problem and that distance and velocity are specified, the acceleration is constant, for that reason we use the equations of uniformly accelerated motion as follows:

v=v_{0}+at\\ x=x_{0}+ v_{0}+\frac{1}{2}at^{2}

If we reorganize the equations, considering that x_{0} and v_{0} are zero because motion starts from rest , we have:

a=\frac{v}{t} \\x=\frac{1}{2}at^{2}  =\frac{1}{2}\frac{v}{t} t^{2}=\frac{1}{2} vt\\t=\frac{2x}{v} \\t=\frac{2*1.80cm}{30.0\frac{cm}{s} } \\t=0.12 s

Finally, the analysis of the result leaves us to understand why the normal heart beat rate varies between 60 and 100 bpm.

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