Answer:
Step-by-step explanation:
First confirm that x = 1 is one of the zeros.
f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26
f(1) = 2 - 14 + 38 - 26
f(1) = -12 + 38 = + 26
f(1) = 26 - 26
f(1) = 0
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next perform a long division
x -1 || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26
2x^3 - 2x^2
===========
-12x^2 + 28x
-12x^2 +12x
==========
26x -26
26x - 26
========
0
Now you can factor 2x^2 - 12x + 26
2(x^2 - 6x + 13)
The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16
So you are going to get a complex result.
x = -(-6) +/- sqrt(-16)
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2
x = 3 +/- 2i
f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)
The zeros are
1
3 +/- 2i
60i
if you go to mode on your calculator then Chang it from "real" to "a+bi" you can solve for unreal numbers. if you press "2nd" then "." it makes an "i"
C=8h+12
the 8h represents the $8 per hour for renting the boat, and h represents however many hours you rent the boat for. The +12 is the initial $12 fee. This is in slope intercept form, just using different variables than y=mx+b
For a binomial experiment in which success is defined to be a particular quality or attribute that interests us, with n=36 and p as 0.23, we can approximate p hat by a normal distribution.
Since n=36 , p=0.23 , thus q= 1-p = 1-0.23=0.77
therefore,
n*p= 36*0.23 =8.28>5
n*q = 36*0.77=27.22>5
and therefore, p hat can be approximated by a normal random variable, because n*p>5 and n*q>5.
The question is incomplete, a possible complete question is:
Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.
Suppose n = 36 and p = 0.23. Can we approximate p hat by a normal distribution? Why? (Use 2 decimal places.)
n*p = ?
n*q = ?
Learn to know more about binomial experiments at
brainly.com/question/1580153
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