Answer:
Quadratic Equation:


From the standard form of a Quadratic Function, we get:

Discriminant:



From the discriminant, we conclude that the equation will have two real solutions.
State that:



By the way, solving the equation given:





Answer:
0.0108 = 1.08% probability that a sophomore non-Chemistry major and then a junior non-Chemistry major are chosen at random.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Probability that a sophomore non-Chemistry major
Out of 92 students, 9 are non-chemistry major sophomores. So

Then a junior non-Chemistry major are chosen at random.
Now, there are 91 students(1 has been chosen), of which 10 are non-chemistry major juniors. So

What is the probability that a sophomore non-Chemistry major and then a junior non-Chemistry major are chosen at random

0.0108 = 1.08% probability that a sophomore non-Chemistry major and then a junior non-Chemistry major are chosen at random.
Answer:√-40=2√-10
Step-by-step explanation:Finding the factors of 40,we have ,√4×√-10=2×√-10=2√-10
Answer:
Step-by-step explanation:
1,2,3,3,3,3,5,6,7,10,10,10,11,11,12,12,13,13,
14,15
Answer:
80 degrees.
Step-by-step explanation: