I'm guessing your problem is this:
y³ - 9y² + y - 9 = 0
right?
In solving this problem, I recommend doing this:
y³ - 9y² + y - 9 = 0
Factor out a y² from the first two numbers in the problem:
y²(y - 9) + (y - 9) = 0
Separate the parentheses which means y - 9 goes on one side. The y² added a one since it came from the + 1 in the middle of expression. When you're separating parentheses like this you just take the outside numbers and combine them together. Since + 1 came from the outside of the (y - 9) and y² also was sitting on the outside of (y - 9) combine them to make y² + 1. Like this:
(y² + 1)(y - 9) = 0
Now separate your two parentheses to two separate problems:
(y² + 1) = 0 and (y - 9) = 0
Now you're y² + 1 will equal:
y² = -1
y = √-1 <-- This number doesn't exist so it will be an imaginary number (i). If you guys didn't learn that in your class I recommend just leaving it as i for that part.
Now solve y - 9 = 0:
y = 9 <-- Since we added nine to both sides to get this.
So you're final answer should be y = i and 9
You have a special type of function, so you have to express the domain and range of each interval. Every aspect of this graph is closed, so you will only use brackets when writing your answers. Hence, you get [30,5] for the first interval, since there is a horizontal distance of 30 and a vertical distance of 5. You repeat the same process for each interval.
C would be the right answer just done the test
Complete question :
It is estimated 28% of all adults in United States invest in stocks and that 85% of U.S. adults have investments in fixed income instruments (savings accounts, bonds, etc.). It is also estimated that 26% of U.S. adults have investments in both stocks and fixed income instruments. (a) What is the probability that a randomly chosen stock investor also invests in fixed income instruments? Round your answer to decimal places. (b) What is the probability that a randomly chosen U.S. adult invests in stocks, given that s/he invests in fixed income instruments?
Answer:
0.929 ; 0.306
Step-by-step explanation:
Using the information:
P(stock) = P(s) = 28% = 0.28
P(fixed income) = P(f) = 0.85
P(stock and fixed income) = p(SnF) = 26%
a) What is the probability that a randomly chosen stock investor also invests in fixed income instruments? Round your answer to decimal places.
P(F|S) = p(FnS) / p(s)
= 0.26 / 0.28
= 0.9285
= 0.929
(b) What is the probability that a randomly chosen U.S. adult invests in stocks, given that s/he invests in fixed income instruments?
P(s|f) = p(SnF) / p(f)
P(S|F) = 0.26 / 0.85 = 0.3058823
P(S¦F) = 0.306 (to 3 decimal places)
