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Setler [38]
3 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]3 years ago
8 0

Answer:

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

Explanation:

yKpoI14uk [10]3 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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Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.
Sergeu [11.5K]
The work done to pull the sled up to the hill is given by
W=Fd
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is W=1900 J and the distance through which the force is applied is d=400 m, so we can calculate the average force by re-arranging the previous equation and by using these data:
F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N
4 0
3 years ago
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st
irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, 31 {\rm {N}/{mm}}. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

\triangle E = 12.79 J

Explanation:

Sprinters' tendons stretch, x_s = 43 mm = 0.043 m

Non athletes' stretch, x_n = 32 mm = 0.032 m

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, E_s = 0.5kx_s^2

Maximum energy stored in the non athletes, E_m = 0.5kx_n^2

Difference in maximum stored energy between the sprinters and the non-athlethes:

\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J

4 0
3 years ago
If a particular atom of an element has five protons a neutral atom of this element would have ________ electrons. A. 0 B. 3 C. 5
Bezzdna [24]
C. 5

a neutral atom has no electrical charge. protons are positive and electrons are negative, they need to be the same to make it a neutral atom.
5 0
2 years ago
Read 2 more answers
If you wanted to soundproof your bedroom, the best material to line the walls would be
Vsevolod [243]

Answer:

a soft foam material because soft materials absorb sound better

6 0
2 years ago
Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0
3 years ago
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