x2=36
We move all terms to the left:
x2-(36)=0
We add all the numbers together, and all the variables
x^2-36=0
a = 1; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·1·(-36)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
x1=−b−Δ√2ax2=−b+Δ√2a
Δ‾‾√=144‾‾‾‾√=12
x1=−b−Δ√2a=−(0)−122∗1=−122=−6
x2=−b+Δ√2a=−(0)+122∗1=122=6
Answer: 0.0241
Step-by-step explanation:
This is solved using the probability distribution formula for random variables where the combination formula for selection is used to determine the probability of these random variables occurring. This formula is denoted by:
P(X=r) = nCr × p^r × q^n-r
Where:
n = number of sampled variable which in this case = 21
r = variable outcome being determined which in this case = 5
p = probability of success of the variable which in this case = 0.31
q= 1- p = 1 - 0.31 = 0.69
P(X=5) = 21C5 × 0.31^5 × 0.69^16
P(X=5) = 0.0241
Answer:The perimeter is 13.6
Step-by-step explanation:
This gives us a perimeter of 13.6
