Answer:
<u>There are:</u>
- 6 red balls - R
- 5 black balls - B
- Total number = 11 balls
<h3>A. Without replacement</h3>
i. <u>Two blacks </u>
ii. <u>The first is black</u>
or, alternatively
- P(BR or BB) = 5/11*6/10 + 2/11 = 3/11 + 2/11 = 5/11
iii. <u>Both are of same colour</u>
- P(BB or RR) = 2/11 + 6/11*5/10 = 2/11 + 3/11 = 5/11
<h3>B. With replacement</h3>
i. <u>Two blacks </u>
- P(BB) = 5/11*5/11 = 25/121
ii. <u>The first is black</u>
or alternatively
- P(BR or BB) = 5/11*6/11 + 25/121 = 30/121 + 25/121 = 55/121 = 5/11
iii. <u>Both are of same colour</u>
- P(BB or RR) = 5/11*5/11 + 6/11*6/11 = 25/121 + 36/121 = 61/121
Just do 30 divided by 7 and put 7 as the denominator on top of the remainder
Answer:
Step-by-step explanation:
(2u+3u)(u+v)-2u+3v
=2u(u+v)+3u(u+v)-2u+3v
=2u^2+2uv+3u^2+3uv-2u+3v
=5u^2+5uv-2u+3v
THe answer is i dont know because you didnt post it :)
Answer:
p=5
Step-by-step explanation:
3p-7+p=13
Add like terms.
3p+p-7=13
4p-7=13
Add 7 to both sides.
4p-7+7=13+7
4p=20
Divide 4 from both sides.
=
p=5
Hope this helps!
If not, I am sorry.
