We are given
Vertical asymptotes:
Firstly, we will factor numerator and denominator
we get
We can see that (x-3) is common in both numerator and denominator
so, we will only set x+3 to 0
and then we can find vertical asymptote
Hole:
We can see that (x-3) is common in both numerator and denominator
so, hole will be at x-3=0
Horizontal asymptote:
We can see that degree of numerator is 2
degree of denominator is also 2
for finding horizontal asymptote, we find ratio of leading coefficients of numerator and denominator
and we get
y=1
now, we can draw graph
Graph:
Answer:
x^3-x^2-22x+10
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that the Acme Company manufactures widgets, which have a mean of 60 ounces and a standard deviation of 7 ounces
We know that 95% of the area lie between -2 and 2 std deviations from the mean.
i.e. Probability for lying in the middle of 95%
Z score 
Between 46 and 74 oz.
b) Between 12 and 57
convert into Z score
P(-6.86<z<-0.43)
=0.5-0.1664=0.3336
c) X<30 gives Z<-4.83
i.e. P(X<30) =0.00
Answer:
a) 0.2416
b) 0.4172
c) 0.0253
Step-by-step explanation:
Since the result of the test should be independent of the time , then the that the test number of times that test proves correct is independent of the days the river is correct .
denoting event a A=the test proves correct and B=the river is polluted
a) the test indicates pollution when
- the river is polluted and the test is correct
- the river is not polluted and the test fails
then
P(test indicates pollution)= P(A)*P(B)+ (1-P(A))*(1-P(B)) = 0.12*0.84+0.88*0.16 = 0.2416
b) according to Bayes
P(A∩B)= P(A/B)*P(B) → P(A/B)=P(A∩B)/P(B)
then
P(pollution exists/test indicates pollution)=P(A∩B)/P(B) = 0.84*0.12 / 0.2416 = 0.4172
c) since
P(test indicates no pollution)= P(A)*(1-P(B))+ (1-P(A))*P(B) = 0.84*0.88+ 0.16*0.12 = 0.7584
the rate of false positives is
P(river is polluted/test indicates no pollution) = 0.12*0.16 / 0.7584 = 0.0253
