q(x)= x 2 −6x+9 x 2 −8x+15 q, left parenthesis, x, right parenthesis, equals, start fraction, x, squared, minus, 8, x, plus, 1
AURORKA [14]
According to the theory of <em>rational</em> functions, there are no <em>vertical</em> asymptotes at the <em>rational</em> function evaluated at x = 3.
<h3>What is the behavior of a functions close to one its vertical asymptotes?</h3>
Herein we know that the <em>rational</em> function is q(x) = (x² - 6 · x + 9) / (x² - 8 · x + 15), there are <em>vertical</em> asymptotes for values of x such that the denominator becomes zero. First, we factor both numerator and denominator of the equation to see <em>evitable</em> and <em>non-evitable</em> discontinuities:
q(x) = (x² - 6 · x + 9) / (x² - 8 · x + 15)
q(x) = [(x - 3)²] / [(x - 3) · (x - 5)]
q(x) = (x - 3) / (x - 5)
There are one <em>evitable</em> discontinuity and one <em>non-evitable</em> discontinuity. According to the theory of <em>rational</em> functions, there are no <em>vertical</em> asymptotes at the <em>rational</em> function evaluated at x = 3.
To learn more on rational functions: brainly.com/question/27914791
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Answer:
to be safe
Step-by-step explanation:
the further you are the better, because we don't know who have the virus from who don't have it it's better to be 6 feet apart frome each other where ever you go.
Answer: 7,280
Step-by-step explanation:
If we take the wall that is feet high and use 1,820 bricks we multiply to get the answer of 7,280. ;) Hope this helps
1.02 is the correct answer i hope this helped<3
Answer:
a. 0.689
b. 0.8
c. 0.427
Step-by-step explanation:
The given scenario indicates hyper-geometric experiment because because successive trials are dependent and probability of success changes on each trial.
The probability mass function for hyper-geometric distribution is
P(X=x)=kCx(N-k)C(n-x)/NCn
where N=4+3+3=10
n=2
k=4
a.
P(X>0)=1-P(X=0)
The probability mass function for hyper-geometric distribution is
P(X=x)=kCx(N-k)C(n-x)/NCn
P(X=0)=4C0(6C2)/10C2=15/45=0.311
P(X>0)=1-P(X=0)=1-0.311=0.689
P(X>0)=0.689
b.
The mean of hyper-geometric distribution is
μx=nk/N
μx=2*4/10=8/10=0.8
c.
The variance of hyper-geometric distribution is
σx²=nk(N-k).(N-n)/N²(N-1)
σx²=2*4(10-4).(10-2)/10²*9
σx²=8*6*8/900=384/900=0.427
