99 POINT QUESTION PLUS BRAINLIEST!!!

1 answer:

7 0

The graphs arey=(1/3)x^2
and y=3
they intersect at x=3
so the limits are x=0 and 3
to rotate it about y=k, minus k from every function

$area=\pi \int\limits^3_0 {( \frac{1}{3}x^2-3)^2-(3-3)^2 } \, dx$
which evaluates to 72/5pi

A is asnwer

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A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If

Blababa [14]

Answer: Our required probability is $\dfrac{1}{2}$

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = $\dfrac{1}{3}$

So, Probability of getting head from fair coin = $\dfrac{1}{2}$

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

$\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}$