Tan(15) = tan(45 - 30)

= [tan(45) - tan(30) ] / [ 1 + tan(30)tan(45)]

= (1 - 1/sqrt(3)) /(1 + 1/sqrt(3))

= (sqrt(3) - 1)/(sqrt(3) + 1)

= (sqrt(3) - 1)^2 /(3 - 1)

= 1/2 [3 + 1 - 2sqrt(3) ]

= (2 - sqrt(3) )

= 0.27 is your answer

6 0

3 years ago

A certain disease occurs most frequently among older women. Of all age groups, women in their 60s have the highest rate of breas

Law Incorporation [45]

Answer:

The probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.

Step-by-step explanation:

Let a set be events that have occurred be denoted as:

S = {A₁, A₂, A₃,..., Aₙ}

The Bayes' theorem states that the conditional probability of an event, say Aₙ given that another event, say X has already occurred is given by:

$P(A_{n}|X)=\frac{P(X|A_{n})P(A_{n})}{\sum\limits^{n}_{i=1}{P(X|A_{i})P(A_{i})}}$

The disease Breast cancer is being studied among women of age 60s.

Denote the events as follows:

B = a women in their 60s has breast cancer

+ = the mammograms detects the breast cancer

The information provided is:

$P(B) = 0.0312\\P(+|B)=0.81\\P(+|B^{c})=0.92$

Compute the value of P (B|+) using the Bayes' theorem as follows:

$P(B|+)=\frac{P(+|B)P(B)}{P(+|B)P(B)+P(+|B^{c})P(B^{c})}$

$=\frac{(0.81\times 0.0312)}{(0.81\times 0.0312)+(0.92\times (1-0.0312)}\\$

$=\frac{0.025272}{0.025272+0.891296}$

$=0.02757\\\approx0.0276$

Thus, the probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.

4 0

2 years ago

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