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labwork [276]
3 years ago
13

In the Atlanta metropolitan area, 723 new cases of invasive cancer of the cervix occurred among white women between 1975 and 198

4, inclusive. An estimated 620,000 white women lived in this area on average during this time period. What is the average annual incidence rate of invasive cervical cancer in this population?
Mathematics
1 answer:
slamgirl [31]3 years ago
3 0

Answer:

1.166 per 1000 of population

Step-by-step explanation:

Given the following :

Newly recorded cases of invasive cervical cancer = 723

Period of time = 1975 - 1984 ( inclusive)

Estimated average Population of white women in the area at the time = 620,000

Average annual. Incidence rate invasive of cervical cancer in the population.

Incidence rate = (new cases / total population) × 10^n

Incidence rate = (723/620000) × 10^n

= 0.0011661

= 1.166 per 1000 of population

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The expression (secx + tanx)2 is the same as _____.
trapecia [35]

<u>Answer:</u>

The expression \bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}

<u>Solution:</u>

From question, given that \bold{(\sec x+\tan x)^{2}}

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On simplication we get

=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}

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=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}

By using the trigonometric identity (a+b)^{2}=a^{2}+2ab+b^{2}

we get 1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}

=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}

=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}

By using the trigonometric identity a^{2}-b^{2}=(a+b)(a-b)  we get 1-\sin ^{2} x=(1+\sin x)(1-\sin x)

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= \frac{1+\sin x}{1-\sin x}

Hence the expression \bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}

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