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Stels [109]
2 years ago
8

Prove that

20" id="TexFormula1" title=" \frac{1 }{ \sin(x) \times \cos(x) } " alt=" \frac{1 }{ \sin(x) \times \cos(x) } " align="absmiddle" class="latex-formula">
equals 2csc2x
​
Mathematics
1 answer:
konstantin123 [22]2 years ago
3 0

Answer:

See below for proof

Step-by-step explanation:

2csc(2x)\\\\2[\frac{1}{sin(2x)}]\\ \\2[\frac{1}{2sin(x)cos(x)}]\\ \\\frac{2}{2sin(x)cos(x)}\\ \\\frac{1}{sin(x)cos(x)}

Thus, the identity is proven

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The sum of two numbers is 15.
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7

Step-by-step explanation:

15-8=7 this is how you answer it

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Lily said that n=3 is a solution of the equation 16+n=18.Is Lily correct ?Explain
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No, Lily is not correct. This is because 16+3 does not equal 18. So to find N, you take 18 and subtract 16. 
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An object has volume 36 cubic inches. A box has side lengths 1 foor by 3 inches by 4 inches.
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<u>4 Objects can fit in the box</u>

Step-by-step explanation:

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How do I write this in sigma notation?
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3 years ago
A multiple-choice examination has 20 questions, each with five possible answers, only one of which is correct. Suppose that one
kari74 [83]

Answer:

The probability that the student answers at least seventeen questions correctly is 8.03\times 10^{-10}.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of correctly answered questions.

It is provided all the questions have five options with only one correct option.

Then the probability of selecting the correct option is,

P(X)=p=\frac{1}{5}=0.20

There are <em>n</em> = 20 question in the exam.

It is also provided that a student taking the examination answers each of the questions with an independent random guess.

Then the random variable can be modeled by the Binomial distribution with parameters <em>n</em> = 20 and <em>p</em> = 0.20.

The probability mass function of <em>X</em> is:

P(X=x)={20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x};\ x =0,1,2,3...

Compute the probability that the student answers at least seventeen questions correctly as follows:

P(X\geq 17)=P (X=17)+P (X=18)+P (X=19)+P (X=20)

=\sum\limits^{20}_{x=17}{{20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x}}\\\\=0.00000000077+0.000000000032+0.00000000000084+0.000000000000042\\\\=0.000000000802882\\\\=8.03\times10^{-10}

Thus, the probability that the student answers at least seventeen questions correctly is 8.03\times 10^{-10}.

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3 years ago
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