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3 years ago

Find the probability of two persons being born on the same day?

Mathematics

2 answers:

Vinil7 [7]3 years ago

3 0

Answer:

1/365

Step-by-step explanation:

The probability of being born any day of the year is 1 or more specifically: 365/365. Since Person B must be born on the same day as Person A their probability is 1/365

guapka [62]3 years ago

3 0

Answer:

Person A can be born on any day of the year since they’re the first person we’re asking. The probability of being born any day of the year is 1 or more specifically: 365/365.

Since Person B must be born on the same day as Person A their probability is 1/365.

We want both of these events to happen so multiply the probabilities:

So you have a 0.27% chance of walking up to a stranger and discovering that their birthday is the same day as yours.

I think you are happy with this answer

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Which is the correct answer

VashaNatasha [74]

Let s denote the number of pages in each issue of the sports magazine.

Let t denote the number of pages in each issue of the technology magazine.

Given that a sports magazine prints 12 issues per year and a technology magazine prints 10 issues per year.The total number of pages in all the issues of the sports magazine for one year is 32 more than the total number of pages in all the issues of the technology magazine for one year.

This can be written in equation as,

$12s=10t+32$

Since, it is given that, "Each issue of the sports magazine has 18 fewer pages than each issue of the technology magazine".

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Hence, Option A is the correct answer.

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aleksandr82 [10.1K]

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Part 1

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Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

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