I would say option 3 hopefully im right!!
Answer:
third option
Step-by-step explanation:
Given
3 ![\left[\begin{array}{ccc}-2&5\\1&0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%265%5C%5C1%260%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Multiply each element in the matrix by 3
= ![\left[\begin{array}{ccc}3(-2)&3(5)\\3(1)&3(0)\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%28-2%29%263%285%29%5C%5C3%281%29%263%280%29%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=
Answer:
A) I) 49 teams
II)63 teams
B) 28 teams
Step-by-step explanation:
A) a team of 5 boys and 6 Girls
=7C5+ 8C6
= 21+28
= 49 teams
a team of 6 boys and 5 girls
= 7C6 + 8C5
= 7+56
= 63 teams
B) one girl is kept constant already and a guy has injury
Girls remaining to choose from 7
Boys remaining to choose from 6
= 1 +7C5 + 6C5
= 1+21+6
= 28 teams
Answer:
72π (in terms of pi)
Step-by-step explanation:
V = π·r²·h
Plug in the values.
V = π·3²·8
Solve.
V = 72π
To leave an equation in terms of pi, just square, multiply, and leave pi at the end.
3² = 9·8 = 72
72π
If you need the answer fully simplified, it is 226.19 (rounded up).
The probability of not drawing C in neither draw is P = 0.5
<h3>
How to get the probability?</h3>
All the cards have the same probability of being drawn, in this case, our set of cards is {F, D, C, G}
The probability of not drawing C is equal to the probability of drawing F, D or G. So we have 3 options out of 4, then the probability is:
p = 3/4.
Now we draw another, this time there are 3 cards, one of these is C, and the other two cards are not C. Then the probability of not drawing C again is equal to 2 over 3.
q = 2/3.
The joint probability (for both of these events to happen) is equal to the product of the individual probabilities:
P = p*q = (3/4)*(2/3) = 0.5
If you want to learn more about probability, you can read:
brainly.com/question/251701