Consider the logarithmic equation f(x) = ln (x + 2) – 1. Round all values to the tenths place.
1 answer:
Hello,
A) x=0==> f(0)=ln(2)-1=-0,30685281944005469058276787854182
y=0==>ln(x+2)-1=0
==>ln(x+2)=1
==>e^ln(x+2)=e^0
==>x+2=0
==>x=-2
B)
dom f(x)={x∈IR | x>2}
x=-2 is the vertical asymptote
c)
A=(-1,-1)
B=(0,ln(2)-1=-0,30685281944005469058276787854182 )
C=(1,ln(3)-1=0,09861228866810969139524523692253)
D=(2,ln(4)-1)=0,38629436111989061883446424291635)
d) see picture
A) x=0==> f(0)=ln(2)-1=-0,30685281944005469058276787854182
y=0==>ln(x+2)-1=0
==>ln(x+2)=1
==>e^ln(x+2)=e^0
==>x+2=0
==>x=-2
B)
dom f(x)={x∈IR | x>2}
x=-2 is the vertical asymptote
c)
A=(-1,-1)
B=(0,ln(2)-1=-0,30685281944005469058276787854182 )
C=(1,ln(3)-1=0,09861228866810969139524523692253)
D=(2,ln(4)-1)=0,38629436111989061883446424291635)
d) see picture
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Answer:
a)
b) We are uncertain
c) It will change significantly
Step-by-step explanation:
a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.
Since we assume that the variances are equal, we use the pooled variance given as
,
where .
The mean difference .
The confidence interval is
b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.
c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.