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2 years ago

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Two unique letters are chosen at random from the alphabet. What is the approximate probability that the first letter chosen is A

?

1 answer:

guapka [62]2 years ago

8 0

Answer:

The approximate probability that the first letter chosen is A = 0.0385....

Step-by-step explanation:

The given statement is two unique letters are chosen at random from the alphabet.

The total number of ways that the two letters would be chosen from alphabets = 650

26*25 = 650

For the first draw there are 26 letters but for the second draw there are only 25 letters left thats why we have multiplied 26 by 25.

The number of ways letters are chosen such that the first one is A, is

1*25 = 25

Thus the probability that the first letter chosen is A:

P(A)25/650 = 0.03846

By rounding off it becomes 0.0385.

Thus the approximate probability that the first letter chosen is A = 0.0385....

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Which ordered pair minimizes the objective function C = 60x + 85y?. (0, 160). (55, 70). (80, 50). (170, 0)

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Step-by-step explanation:

The given objective function is $C = 60x + 85y$

To check that which ordered pair is minimizing the objective function-we will substitute all the ordered pair one by one in the given function and we will calculate the value of C. The ordered pair which will give the minimum value will be the answer.

For (0, 160)

$C=60\times 0+85\times 160=13600$

For (55, 70)

$C=60\times 55+85\times 70=9250$

For (80, 50)

$C=60\times 80+85\times 50=9050$

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Hence the ordered pair which minimizes the objective function is (80, 50)

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2 years ago

Read 2 more answers

Adam must fly home to city A from a business meeting in city B. One flight option flies directly to city A from B, a distance o

g100num [7]

Answer:

The value is $k =109.6 \ miles$

Step-by-step explanation:

The diagram illustrating the question is shown on the first uploaded image

From the question we are told that

The distance from city A to B is AB = 467.3 miles

The bearing from B to C is $\theta_{BC} = N 28.7E$

The bearing from B to A is $\theta_{BA} = N 60.7E$

The bearing from A to B is $\theta_{AB} = S60.7W$

The bearing from A to C is $\theta_{AC} = S79.1W$

Generally from the diagram

$\theta_A = 180 - 60.7 -79.1$

=> $\theta_A = 40.2 ^o$

Also

$\theta_B = 32^o$

and

$\theta_C = 180 - (\theta_A +\theta_B )$

=> $\theta_C = 180 - (40.2 + 32 )$

=> $\theta_C = 107.8 ^o$

Generally according to Sine Rule

$\frac{BC}{sin (\theta_A)} = \frac{CA}{sin (\theta_B)} =\frac{AB}{sin (\theta_C)}$

=> $\frac{BC}{sin (40.2)} = \frac{CA}{sin (32)} =\frac{467.3 }{sin (107.8)}$

So

$\frac{BC}{sin (40.2)} = \frac{467.3 }{sin (107.8)}$

=> $BC = 316.8 \ miles$

Also

$\frac{CA}{sin (32)} = \frac{467.3 }{sin (107.8)}$

$CA = 260 .1 \ miles$

Generally the additional flyer miles that Adam will receive if he takes the connecting flight rather than the direct flight is mathematically represented as

$k = [CA +BC] - AB$

=> $k = [260 .1 +316.8]- 467.3$

=> $k =109.6 \ miles$

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2 years ago