18. If f(x)=[xsin πx] {where [x] denotes greatest integer function}, then f(x) is:
since x denotes the greatest integers which could the negative or the positive values, also x has a domain of all real numbers, and has no discontinuous point, then x is continuous in (-1,0).
Answer: B]
20. Given that g(x)=1/(x^2+x-1) and f(x)=1/(x-3), then to evaluate the discontinuous point in g(f(x)) we consider the denominator of g(x) and f(x). g(x) has no discontinuous point while f(x) is continuous at all points but x=3. Hence we shall say that g(f(x)) will also be discontinuous at x=3. Hence the answer is:
C] 3
21. Given that f(x)=[tan² x] where [.] is greatest integer function, from this we can see that tan x is continuous at all points apart from the point 180x+90, where x=0,1,2,3....
This implies that since some points are not continuous, then the limit does not exist.
Answer is:
A]
4x-6= 22
Move -6 to the other side.
Sign changes from -6 to +6.
4x-6+6= 22+6
4x= 22+6
4x= 28
Divide by 4 for both sides
4x/4= 28/4
x= 7
Answer : x= 7
Answer:
Answer:
a).
The amount spent on school materials for each term of all ST201students
b).
a).
It is not a random sample. This looks like a convenience sampling and there is sampling bias. This sample is not representative of the entire population. Since it is not a random sample it is not appropriate to generalize the results to all students.
b).
The sample size is 80 which is greater than 30. It is large enough to assume normal distribution according to central limit theorem.
c).
mean: $617
z critical value at 95%: 1.96
standard error = σ/sqrt(n) =500/sqrt(80) = 55.9017
lower limit= mean-1.96*se = 617-1.96*55.9017=507.43
upper limit= mean+1.96*se = 617+1.96*55.9017=726.57
d).
The amount spent on school materials for each term for the 80 ST201students is $617. We are 95% confident that amount spent on school materials for each term of all ST201students falls in the interval ($507.43, $726.57).
Step-by-step explanation:
