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KiRa [710]
2 years ago
11

Determine an exact value for sin _ 3π/4-tan 5π/6

Mathematics
1 answer:
iris [78.8K]2 years ago
6 0

Answer:

\frac{3\sqrt{2}+2\sqrt{3}}{6} or \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}

Step-by-step explanation:

\sin\frac{3\pi}{4}-\tan\frac{5\pi}{6}\\\\\sin\frac{3\pi}{4}-\frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}}\\ \\\frac{\sqrt{2}}{2}-\frac{\frac{1}{2} }{-\frac{\sqrt{3}}{2}}\\ \\\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{3}}\\ \\\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}\\ \\\frac{3\sqrt{2}}{6}+\frac{2\sqrt{3}}{6}\\ \\\frac{3\sqrt{2}+2\sqrt{3}}{6}

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Answer:

The value of x are all real number on x-axis the number that falls under x

example of the value of x are in this equation

y=2+5x

2x+6y=60

to find x

let make the first one equation 1 and the second equation 2

now, put equation 1 in equation 2

2x+6(2+5x)=92

open the by multiplying every thing in the bracket by 6

2x+12+30x=92

combine the like terms

2x+30x=69212

32x=80

now divide via by 32

32x/32=80/32

x=5/2

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3 years ago
A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f
Mrrafil [7]

Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

3) The contestant gets the question 2 and gets the question 1 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

3 0
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Let x represent the number of minutes for which her call lasted.

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This means that the cost of x minutes of long distance call is

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0.06x = 30 - 28.44

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