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Naddik [55]
3 years ago
6

What is the probability that a randomly selected person will have a birthday in march? assume that this person was not born in a

leap year. express your answer as a simplified fraction or a decimal rounded to four decimal places?
Mathematics
2 answers:
san4es73 [151]3 years ago
8 0

Answer:

0.0849

Step-by-step explanation:

Let's consider the event: a birthday takes place in March. The probability (P) of such event is:

P = \frac{favorable\ cases }{possible\ cases}

The favorable cases are 31 because March has 31 days.

The possible cases are 365 because there are 365 days in a year.

The probability of a birthday being in March is:

P=\frac{31}{365} =0.0849

ivann1987 [24]3 years ago
5 0

there are 31 days in August

 365 days in a year

31/365 = 0.0849 probability

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d

Step-by-step explanation:

Slope intercept form: y = mx + b

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8 0
2 years ago
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Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

Let, \text{sin}^{-1}(\frac{x}{6}) = y

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\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx} ---------(1)

\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

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cos(y) = \sqrt{1-\text{sin}^{2}(y) }

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Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

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