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ale4655 [162]
3 years ago
10

A pet store has 8 ​puppies, including 3 ​poodles, 2 ​terriers, and 3 retrievers. If Rebecka selects one puppy at​ random, the pe

t store replaces the puppy with a puppy of the same​ breed, then Aaron chooses a puppy at random. Find the probability that they both select a poodle.
Mathematics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

9 / 64

Step-by-step explanation:

- In this task you have 2 events and you are looking for a joint probability. The first event is "Rebecca chooses a poodle". The probability of this event is:

                    P ( Rebecca chooses a poodle ) = 3 / 8

- because among 8 dogs there are 3 poodles.

- The second event is "Aaron selects a poodle".

This event has a probability of  that is equivalent to previous selection:

                   P ( Aaron chooses a poodle ) = 3 / 8

- Because after Rebecca's choice the chosen poodle is replaced with the poodle; hence, there are 8 pets in total and among them there are 3 poodles.

- To calculate probability of both events ("Rebeca selects a poodle and Aaron selects a poodle") with replacement you have to multiply both calculated probabilities - condition of independent events :

     P ( Aaron and Rebecca both select poodle ) = 3 / 8 * 3 / 8

                                                                                = 9 / 64

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forsale [732]

① 15² = 4² + 2(32) x
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② 105 = 45 + 5x
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Find a1, for the arithmetic series<br> with S14 = - 420 and d = -6.
faust18 [17]

\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} n = 14\\ d= -6 \end{cases} \\\\\\ a_{14}=a_1+(14-1)(-6)\implies a_{14}=a_1+(13)(-6)\implies a_{14}=a_1-78 \\\\[-0.35em] ~\dotfill

\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ \cline{1-1} n= 14\\ S_{14}=-420\\ a_{14}=a_1-78 \end{cases}\implies S_{14}=\cfrac{n(a_1+a_{14})}{2} \\\\\\ -420=\cfrac{14[a_1+(a_1-78)]}{2}\implies -420=7(2a_1-78)\implies \cfrac{-420}{7}=2a_1-78 \\\\\\ -60=2a_1-78\implies 18=2a_1\implies \cfrac{18}{2}=a_1\implies 9=a_1

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Step-by-step explanation:

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