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3 years ago

8

How could you explain the sine and cosine relationships to someone who doesn't know about the unit circle? What visuals could yo

u use to help explain the idea of how these two measures are related?

1 answer:

sukhopar [10]3 years ago

8 0

If you have a unit circle, it means that you have a circle of radius r = 1 unit. So if we consider the scalene triangle of the figure, its vertical side is the sine of the theta angle and its horizontal side is the cosine of the theta angle. Then to find the hypotenuse of the triangle you use the Pythagorean theorem, thus relating the sine and cosine measures. I attach the solution.

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Answer:

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Step-by-step explanation:

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3 years ago

What is the next number on sequence 2 7 8 3 12 9

Kamila [148]

The term of this sequence is:
<span>-(17/30)n^5+(113/12)n^4-(173/3)n^3+(1915/12)n^2-(5813/30)n+85 </span>
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3 years ago

You’ve bought a half-dozen (six) eggs from the store but you forgot to check them first! The probability that no eggs are broken

GREYUIT [131]

Answer:

a)

$P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4$

$P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960$

$P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633$

$P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359$

$P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044$

$P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003$

$P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001$

b) 56.77% probability that an even number of eggs is broken.

c)

Expectation: 0.8496

Variance: 0.7293

Step-by-step explanation:

For each egg, there are only two possible outcomes. Either it is broken, or it is not. The probability of an egg being broken is independent from other eggs. So we use the binomial probability distribution 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.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

There are 6 eggs

So $n = 6$

The probability that no eggs are broken is 0.4.

This means that $P(X = 0) = 0.4$ . So

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

$P(X = 0) = C_{6,0}.p^{0}.(1-p)^{6}$

$(1 - p)^{6} = 0.4$

Taking the sixth root from both sides of the equality

$(1 - p) = 0.8584$

p = 0.1416

Each egg has a 0.1416 probability of being broken

(a) Write out the pmf of X.

Probability of each value, from 0 to 6

$P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4$

$P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960$

$P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633$

$P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359$

$P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044$

$P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003$

$P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001$

(b) Compute the probability that an even number of eggs is broken.

0, 2, 4 or 6

$P = P(X = 0) + P(X = 2) + P(X = 4) + P(X = 6) = 0.4 + 0.1633 + 0.0044 + 0.00001 = 0.5677$

56.77% probability that an even number of eggs is broken.

(c) Compute the expectation and variance of X.

Expectation:

$E(X) = np = 6*0.1416 = 0.8496$

Variance:

$V(X) = np(1-p) = 6*0.1416*0.8584 = 0.7293$

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What is 0.000000783 written in scientific notation?

UNO [17]

Answer:

7.83 * 10^-7.

Step-by-step explanation:

Because the number is < 0 the exponent will be negative. There are 7 digits after the decimal point up to and including the 7, so the exponent is -7.

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4 years ago

Cody is carrying seven pages of math homework and three pages of English homework. A gust of wind blows the pages out of his han

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Answer:

1/20

Step-by-step explanation:

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2 years ago

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