C.x=-7.7 and x=-1.3 should be the answer
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
I have no idea I just need more points so I can ask more questions
Answer:
I would say the answer is b
Step-by-step explanation:
Answer:
Therefore,
Step-by-step explanation:
Given:
![A=\left[\begin{array}{ccc}3&6&9\\2&4&8\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%269%5C%5C2%264%268%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To Find:
a₂₁ = ?
Solution:
Let,
![A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26a_%7B12%7D%26a_%7B13%7D%5C%5Ca_%7B21%7D%26a_%7B22%7D%26a_%7B23%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
We require ' a₂₁ ' i.e Second Row First Column Element
So on Comparing we get
∴
Therefore,
