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3 years ago

5a+5b+5c=-20 4a+3b+3c=-6 -4a+3b+3c=9

1 answer:

4 0

For 5a+5b+5c=-20

A= -4 - b - c

B= -4 - a - c

C= -4 - a - b

———————————

4a+3b+3c=-6

A= -3/2 - 3b/4 - 3c/4

B= -2 - 4a/3 - c

C= -2 - 4a/3 - b

———————————

-4a+3b+3c=9

A= -9/4 + 3b/4 + 3c/4

B= 3 + 4a/3 - c

C= 3 + 4a/3 - b

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Step-by-step explanation:

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A center for medical services reported that there were 295,000 appeals for hospitalization and other services. For this group, 4

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Answer:

a) 0.0025 = 0.25% probability that none of the appeals will be successful.

b) 0.0207 = 2.07% probability that exactly one of the appeals will be successful.

c) 0.9768 = 97.68% probability that at least two of the appeals will be successful

Step-by-step explanation:

For each appeal, there are only two possible outcomes. Either it is succesful, or it is not. The probability of an appeal being succesful is independent of other appeals, 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.

45% of first-round appeals were successful.

This means that $p = 0.45$

Suppose 10 first-round appeals have just been received by a Medicare appeals office.

This means that $n = 10$

(a) Compute the probability that none of the appeals will be successful.

This is P(X = 0).

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

$P(X = 0) = C_{10,0}.(0.45)^{0}.(0.55)^{10} = 0.0025$

0.0025 = 0.25% probability that none of the appeals will be successful.

(b) Compute the probability that exactly one of the appeals will be successful.

This is P(X = 1).

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

$P(X = 1) = C_{10,1}.(0.45)^{1}.(0.55)^{9} = 0.0207$

0.0207 = 2.07% probability that exactly one of the appeals will be successful.

(c) What is the probability that at least two of the appeals will be successful

This is $P(X \geq 2)$

Either less than two appeals are succesful, or at least two are. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0207 = 0.0232$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0232 = 0.9768$

0.9768 = 97.68% probability that at least two of the appeals will be successful

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Tan 3A in terms of tan

Zanzabum

Here's the sum rule for the tangent function:

$\tan(a+b)=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$

In the special case a=b, this becomes the double angle formula:

$\tan(a+a)=\tan(2a)=\dfrac{\tan(a)+\tan(a)}{1-\tan(a)\tan(a)}=\dfrac{2\tan(a)}{1-\tan^2(a)}$

In your case, you case use the sum rule once:

$\tan(3a)=\tan(2a+a)=\dfrac{\tan(2a)+\tan(a)}{1-\tan(2a)\tan(a)}$

And use it again, in the special case of the double angle:

$\dfrac{\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a)}{1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a)}$

We can obvisouly simplify this expression a lot: let's deal with the numerator and denominator separately: the numerator is

$\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a) = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}$

and the denominator is

$1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a) = \dfrac{1-\tan^2(a)-2\tan^2(a)}{1-\tan^2(a)} = \dfrac{1-3\tan^2(a)}{1-\tan^2(a)}$

So, the fraction is

$\dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}\cdot \dfrac{1-\tan^2(a)}{1-3\tan^2(a)} = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-3\tan^2(a)}$

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4 years ago