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evablogger [386]
3 years ago
9

Denis examined the records of the clients who had joined his gym more than six months ago, and he found these statistics: \begin

{aligned} &P(\text{joined in January})=0.12 \\\\ &P(\text{member for over 6 months})=0.5 \\\\ &P(\text{over 6 months and January})=0.024 \end{aligned} ​ P(joined in January)=0.12 P(member for over 6 months)=0.5 P(over 6 months and January)=0.024 ​ Find the probability that a client remained a member for more than 666 months, given that the client joined in January.
Mathematics
1 answer:
drek231 [11]3 years ago
8 0

Answer:

.2

Step-by-step explanation:

This is a conditional probability question. So it's asking you to find the probability that a client remained a member for more than 6 months, given that the client joined in January - which is formatted as P = (.5 | .12). You would then divide the chance of being over 6 months AND in January over the chance of being a member for over six months. ( .024 / .12) There, you would get .2 as your answer.

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Verify cot x sec^4x=cotx +2tanx +tan^3x
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Answer:

See explanation

Step-by-step explanation:

We want to verify that:

\cot(x)  \:  { \sec}^{4} x =  \cot(x) + 2 \tan(x)   +  { \tan}^{3} x

Verifying from left, we have

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \: ( 1 +  { \tan}^{2} x )^{2}

Expand the perfect square in the right:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \: ( 1 +  { 2\tan}^{2} x  + { \tan}^{4} x)

We expand to get:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \:   +  \cot(x){ 2\tan}^{2} x  +\cot(x) { \tan}^{4} x

We simplify to get:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \:   +  2 \frac{ \cos(x) }{\sin(x) ) }  \times  \frac{{ \sin}^{2} x}{{ \cos}^{2} x}   +\frac{ \cos(x) }{\sin(x) ) }  \times  \frac{{ \sin}^{4} x}{{ \cos}^{4} x}

Cancel common factors:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \:   +  2 \frac{{ \sin}x}{{ \cos}x}   +\frac{{ \sin}^{3} x}{{ \cos}^{3} x}

This finally gives:

\cot(x)  \:  { \sec}^{4} x =  \cot(x) + 2 \tan(x)   +  { \tan}^{3} x

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