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3 years ago

5/√2+9/√8-2√50+√32 rationalise the denominator and simplify

Mathematics

1 answer:

Alik [6]3 years ago

7 0

Answer: $\displaystyle\ \Large \boldsymbol{ -\frac{5\cdot \sqrt{2} }{4}}$

Step-by-step explanation: $\displaystyle\ \Large \boldsymbol{} \frac{5}{\!\!\sqrt{2} } +\frac{9}{\!\!\sqrt{8} } -2\sqrt{50} +\sqrt{32} =\frac{1}{\!\!\sqrt{2} } \bigg(5+\frac{9}{\sqrt{4} } \bigg)+\sqrt{2} (\sqrt{16} -2\sqrt{25} ) \\\\\\\\\frac{1}{\sqrt{2} } (5+4,5)+\sqrt{2} (4-10)=\frac{9,5}{\sqrt{2} } -6\sqrt{2} =\frac{9,5-12}{\sqrt{2} } = \\\\\\-\frac{5\cdot \sqrt{2} }{4}$

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Add 55 to both sides.

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Simplify 8+58+5 to 1313.

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a = 36, b = 54

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Since the terms are in arithmetic progression then there is a common difference d between consecutive terms , that is

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Thus a = 36 and b = 54

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b) The probability of testing positive is 6.4%

c) The probability of not using steroids, given that the test is negative is 99.47%

d) No, they are not statistically indepent.

e) The probability that the athlete will either use steroids or test positive is 6.9%

Step-by-step explanation:

Let A be the event that the test result is positive and B the event that the athlete uses Steroids. We are given the following

$P(A|B) = 90%, P(A|B^c) = 2%, P(B) = 5%$

From which we deduce that

$P(A^c|B) = 10%, P(B^c) = 95%$

a) We are asked for the probability $P(A^c\cap B)$ . REcall the conditional probability formula that, given two events C,D the conditional probability $P(C|D) = \frac{P(C\cap D)}{P(D)}$ . Then we have that

$P(A^c\cap B) = P(A^c|B)P(B) = 10\% \cdot 5\%=0.5\%$ .

b) We are asked for the probability P(A). We can use the fact that given two mutually exclusive events(that is, whose intersection is empty) A,B the probability P(C) of an event is given by $P(C) = P(C|A)P(A)+P(C|B)P(B)$ . Then

$P(A) = P(A|B)P(B)+P(A|B^c)P(B^c) = 90\%\cdot5\% + 2\% \cdot 95% = 6.4\%$

c) We are asked for the probability P(B^c|A^c). Recall that $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ . Then

$P(B^c|A^c) = \frac{P(A^c|B^c)P(B^c)}{P(A^c)}= \frac{P(A^c|B^c)P(B^c)}{1-P(A)}= \frac{(1-P(A|B^c))P(B^c)}{1-P(A)}=\frac{98\%\cdot 95\%}{1-6.4\%}= 99.47\%$

d) We say that two events A,B are statistically indepent if P(A|B) = P(A). Note that from point B the probability of testing negative is 1- 6.4% = 93.6%. Since 93.6% is different from 99.47% this means that testing positive and using steroids are not statistically independent.

e) We are asked for the probability $P(A\cup B)$ . We use the following