Evaluate this piecewise function at x = -8 and x = 4.
1 answer:
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Consider the functions m and n:
i)
for x≠1.
the domain of m is all the real numbers except 1.
ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial.
The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:
x-3=c, then x=3+c.
so if we want the function to produce c, we just set x=3+c, check:
n(3+c)=(3+c)-3=c.
This means that the Range of n is all real numbers.
What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.
so let n(x)=a, consider m(a):
m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4
This means that the domain of the composed function, m(n(x)) = say f(x),
is all Real numbers except 4.
Check: letting x be 4, would mean n(4)=4-3=1
which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.
Answer: Any function with domain R-{4}, has the same domain of (mon)(x)
i)
the domain of m is all the real numbers except 1.
ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial.
The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:
x-3=c, then x=3+c.
so if we want the function to produce c, we just set x=3+c, check:
n(3+c)=(3+c)-3=c.
This means that the Range of n is all real numbers.
What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.
so let n(x)=a, consider m(a):
m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4
This means that the domain of the composed function, m(n(x)) = say f(x),
is all Real numbers except 4.
Check: letting x be 4, would mean n(4)=4-3=1
which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.
Answer: Any function with domain R-{4}, has the same domain of (mon)(x)
Answer:
The calculated value z = 9.4451 > 1.96 at 5% level of significance.
Null hypothesis is rejected at 5% level of significance
yes there is difference in the distribution of types of cell phones for the teachers in 2018 at a 5% level of significance
Step-by-step explanation:
<u>Explanation</u>:-
<u>Step:- (1)</u>
The results of a 2012 Pew Foundation survey of high school and middle school teachers is given in the pie chart.
A student asked a random sample of teachers in 2018 and found 165 had smart-phones, 80 had a cell phone
The first sample proportion
A student asked a random sample of teachers in 2018 and found 165 had smart-phones,5 had no cell phone
The second sample proportion
<u>Step :-(ii)</u>
<u>Null hypothesis :H₀</u>: Assume that there is no difference in the distribution of types of cell phones for the teachers in 2018
H₀ : p₁ = p₂
<u>Alternative hypothesis :H₁</u>
H₁ : p₁ ≠ p₂
<u>Level of significance : ∝=0.05</u>
The tabulated value z=1.96
<u>Step:-(iii)</u>
The test statistic
where
q = 1-p
In given data n₁ = n₂ = n
on calculation , we get p = 0.2655
q =1-p = 1-0.2655
q = 0.7345
Z = 9.4451
The calculated value z = 9.4451 > 1.96 at 5% level of significance.
<u>Conclusion:</u>-
Null hypothesis is rejected at 5% level of significance
yes there is difference in the distribution of types of cell phones for the teachers in 2018 at a 5% level of significance