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AfilCa [17]
3 years ago
6

(adapted from Ross, 2.31) Three countries (the Land of Fire, the Land of Wind, and the Land of Earth) each make a 3 person team.

Each team consists of three ninjas, each of a different rank: one genin junior ninja, one journeyman ninja, and one elite ninja.
(a) If a ninja is chosen at random from each of the three different countries, what is the probability of selecting a complete team? (A complete team here is a group consisting of three ninjas, each ninja of a different rank).

(b) If a ninja is chosen at random from each of the three different countries, what is the probability that all 3 ninjas selected have the same rank?
Mathematics
2 answers:
DaniilM [7]3 years ago
4 0

Answer:

The answer is "\frac{2}{9} \  and \ \frac{1}{9}"

Step-by-step explanation:

In point a:

The requires  1 genin, 1 chunin , and 1 jonin to shape a complete team but we all recognize that each nation's team is comprised of 1 genin, 1 chunin, and 1 jonin.

They can now pick 1 genin from a certain matter of national with the value:

\frac{1}{\binom{3}{1}}=\frac{1}{3} .

They can pick 1 Chunin form of the matter of national with the value:

\frac{1}{\binom{3}{1}}=\frac{1}{3} .

They have the option to pick 1 join from of the country team with such a probability: \frac{1}{\binom{3}{1}}=\frac{1}{3}

And we can make the country teams: 3! = 6 different forms. Its chances of choosing a team full in the process described also are:

6 \times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{2}{9}.

In point b:

In this scenario, one of the 3 professional sides can either choose 3 genins or 3 chunines or 3 joniners. So, that we can form three groups that contain the same ninjas (either 3 genin or 3 chunin or 3 jonin).

Its likelihood that even a specific nation team ninja would be chosen is now: \frac{1}{\binom{3}{1}}=\frac{1}{3}

Its odds of choosing the same rank ninja in such a different country team are: \frac{1}{\binom{3}{1}}=\frac{1}{3}

The likelihood of choosing the same level Ninja from the residual matter of national is: \frac{1}{\binom{3}{1}}=\frac{1}{3} Therefore, all 3 selected ninjas are likely the same grade: 3\times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{1}{9}

Elenna [48]3 years ago
3 0

Answer: ITS 2 WEEKS AGO IT DOSE T MATTER IF IM RIGHT OR NOT HEHHEHEJSSHAGAGAHAHHAHAA

Step-by-step explanation:

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(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

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\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

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In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

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|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

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\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

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