Answer:
80 units squared
Step-by-step explanation:
well first you get the area of the big rectangle, 8x12=96
then you need to get the area of the light sqaure, because there is the same amount of ticks 3 times, the equal lengths are 12/3= 4 units each so the dimensions of the sqaure are 4 by 4 4x4=16 then you do 96-16= 80 units squared.
Answer:
6x -3x^4
Step-by-step explanation:
Eliminate parentheses. Since the multiplying coefficient is +1, the parentheses can simply be dropped:
= 3x +2x^4 +3x -5x^4
Add like terms (ones with the same exponent on the variable).
= 3x +3x +2x^4 -5x^4 . . . . . . . group like terms together
= (3+3)x +(2-5)x^4
= 6x -3x^4
4s = 3r
<span>s = 3/4*r </span>
<span>plug this into the second eqution and solve for the other variable: </span>
<span>2r +5(3/4)r = 23 </span>
<span>23/4*r = 23 </span>
<span>r = 4 </span>
<span>plug this r into the above equation we solved: </span>
<span>s = 3/4*r </span>
<span>s = 3/4*4 </span>
<span>s = 3 </span>
<span>system is: </span>
<span>{r=4, s=3}</span>
The probability of event A and B to both occur is denoted as P(A ∩ B) = P(A) P(B|A). It is the probability that Event A occurs times the probability that Event B occurs, given that Event A has occurred.
So, to find the probability that you will be assigned a poem by Shakespeare and by Tennyson, let Event A = the event that a Shakespeare poem will be assigned to you; and let Event B = the event that the second poem that will be assigned to you will be by Tennyson.
At first, there are a total of 13 poems that would be randomly assigned in your class. There are 4 poems by Shakespeare, thus P(A) is 4/13.
After the first selection, there would be 13 poems left. Therefore, P(B|A) = 2/12
Based on the rule of multiplication,
P(A ∩ B) = P(A) P(B|A)P(A ∩ B) = 4/13 * 2/12
P(A ∩ B) = 8/156
P(A ∩ B) = 2/39
The probability that you will be assigned a poem by Shakespeare, then a poem by Tennyson is 2/39 or 5.13%.
Answer:
-4
Step-by-step explanation:
i just did it on edge and got it right
