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drek231 [11]
4 years ago
13

You want to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter. What is the probability

that a randomly chosen code starts with M and ends with E?
Mathematics
1 answer:
Feliz [49]4 years ago
4 0

Answer:

The probability that a randomly chosen code starts with M and ends with E is 0.05 ....

Step-by-step explanation:

According to the given statement we have to make five letter code from  A, F, E, R, and M without repeating any letter. We have to find that what is probability that a randomly chosen code starts with M and ends with E.

Thus the probability of picking the first letter M = 1/5

After that we require the sequence (not E, not E, not E) which is equal to:

= 3/4 * 2/3* *1/2

= 1/4

Now multiply 1/5 and 1/4

1/5 * 1/4

= 1/20

= 0.05

Therefore the probability that a randomly chosen code starts with M and ends with E is 0.05 ....

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Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

T_1cos\theta_1-T_2cos\theta_2=0           (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

T_1sin\theta_1+T_2sin\theta_2-W=0               (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2

Then, you sum the equations:

T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2        (3)

Next, you use the following identity:

sin^2\theta+cos^2\theta=1

and you obtain in the equation (3):

T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°

With this values you can calculate the value of the another angle, by using the equation (1):

\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°

Now, you can calculate the length of each cable by using the information about the width of the football field. You use the following trigonometric relation:

l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

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