Answer:
x=2, y=3
Step-by-step explanation:
This is a <u>system of equations</u>.
To solve a system of equations, we use the substitution method.
We can plug in the first equation,
, into the value for <em>x</em> in the second equation. Now, the second equation looks like this:

We simplify and solve for <em>y</em>:
→ 
→ 
→ 
Now, we plug in the value for <em>y</em>, which is 3, into either equation. I will just use the first equation:
→ 
→ 
Let me know if my explanation is unclear.
Answer:
P(C4) = 0.0711
Step-by-step explanation:
consider the first draw = 15/23 since it cannot be a blue ball
The second draw = 21/29 since 6 more red balls will be added after the draw since a blue ball cannot be drawn
the third draw = 27/35 since 6 more red balls will be added after each draw since a blue ball cannot be drawn
therefore the total number of red balls will be = 15 + 6 + 6 + 6 = 33 red balls after the 4th draw. the total ball now in the urn= 33 red + 4 blue = 41
Hence the probability of drawing a blue ball at the fourth draw after drawing red balls at the previous attempts = 8/41
P(C4) = P ( fourth ball is blue ) * P( first ball red)*P(second ball red) *P(third ball red )
= (8/41) * (15/23) * (21/29)* (27/35) = 0.0711
The top right is the answer. The base should be square with lengths of 8 and the height is 10.
Answer:
x ≥ 12
Hope that helps!
Step-by-step explanation:
Answer:
I cannot answer all of this, so I will only answer what I can:
Q1: 13
Q3: 22.5
IQR: 13.5
IQR(1.5): 20.25
Q3 + IQR(1.5) = 43
Q1 - IQR(1.5) = -7.25
I don't know if my calculations are all right but I hope this helps! :)