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Arturiano [62]
3 years ago
9

Find sin(2x), cos(2x) and tan (2x) from the given information. cot(x)=2/3, x in quadrant I

Mathematics
1 answer:
Usimov [2.4K]3 years ago
4 0

Answer:

1.) 6√13/13

2.) 4√13/13

3.) 3

Step-by-step explanation:

given that cot(x) = 2/3

But cot(x) = 1/tan(x)

Substitutes 1/tan(x) for cot(x)

1/tan(x) = 2/3

Reciprocate both sides

Tan(x) = 3/2 = opposite/adjacent

Use pythagorean theorem to find the hypothenus.

Hypothenus = sqrt ( 3^2 + 2^2 )

Hypothenus = sqrt ( 9 + 4 )

Hypothenus = sqrt (13)

Hypothenus = √13

1.) Sin(x) = opposite/hypothenus = 3/√13

Rationalise

3/√13 × √13/√13

3√13/13

Sin(2x) = 2 × 3√13/13

Sin(2x) = 6√13/13

2.) Cos( x ) = adjacent/hypothenus

Cos (x) = 2/√13

Rationalise

2/√13 × √13 /√13

2√13/13

Cos(2x) = 2 × 2√13/13

Cos(2x) = 4√13/13

3.) Tan (x) = 3/2

tan (2x) = 2 × 3/2

Tan(2x) = 3

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When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal. During one
Triss [41]

Answer:

0.82 = 82% probability that at least one party will not order drinks

Step-by-step explanation:

For each party, there are only two possible outcomes. Either they will order drinks with their meal, or they will not. The probability of a party ordering drinks with their meal is independent of other parties. So the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal.

This means that p = 0.75

During one hour, Ryan served 6 parties. Assuming that each party is equally likely to order drinks, what is the probability that at least one party will not order drinks?

6 parties, so n = 6.

Either all parties will order drinks, or at least one will not. The sum of the probabilities of these events is decimal 1. So

P(X = 6) + P(X < 6) = 1

We want P(X < 6). So

P(X < 6) = 1 - P(X = 6)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.18

P(X < 6) = 1 - P(X = 6) = 1 - 0.18 = 0.82

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5 0
3 years ago
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oee [108]

Option C:

Area of the remaining paper = (3x – 4)(3x + 4) square centimeter

Solution:

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Area of the square corner removed = 16 sq. cm

Let us find the area of the remaining paper.

Area of the remaining paper = Area of the square paper – Area of the corner

Area of the remaining = 9x^2-16

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Using algebraic formula: a^2-b^2=(a+b)(a-b)

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Area of the remaining paper = (3x – 4)(3x + 4) square centimeter

Hence (3x – 4)(3x + 4)  represents area of the remaining paper in square centimeters.

8 0
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Answer:

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Answer:

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