|x - 3| - 1 = 5
x - 3 - 1 = 5
x - 4 = 5
x = 9
-(x - 3) - 1 = 5
-x + 3 - 1 = 5
-x + 2 = 5
-x = 3
x = - 3
x = -3 , 9 matched the answers on the number line
Answer is B) |x - 3| - 1 = 5
48. B and 47 is out of c and b
Answer:
(1238.845 ;1285.376)
Step-by-step explanation:
Conditions for constructing a confidence interval :
Data must be random
Distribution should be normal and independent ;
Based on the conditions above ; data meets initial conditions ;
C. I = sample mean ± margin of error
Given the data :
1241 1210 1267 1314 1211 1299 1246 1280 1291
Mean, xbar = Σx / n = 11359 / 9 = 1262.11
The standard deviation, s = [√Σ(x - xbar)²/n - 1]
Using a calculator ; s = 37.525
The confidence interval :
C.I = xbar ± [Tcritical * s/√n]
Tcritical(0.10 ; df = n - 1 = 9 - 1 = 8)
Tcritical at 90% = 1.860
C. I = 1262.11 ± [1.860 * 37.525/√9]
C.I = 1262.11 ± 23.266
(1238.845 ;1285.376)
± 23.266
The margin of error :
[Tcritical * s/√n]
[1.860 * 37.525/√9]
C.I = ± 23.266
Answer:
cl + 1/(N+1)
Step-by-step explanation:
If we assume that the Nth harmonic number is cl. Then we are assuming that 1+1/2+1/3+1/4+...+1/N=cl
And we know that the (N+1)th harmonic number can be found by doing
1+1/2+1/3+1/4+...+1/N+1/(N+1)
=cl + 1/(N+1)
The (N+1)th harmonic number is cl + 1/(N+1) given that the Nth term is cl
Other way to see the answers:
Maybe you want to write it as a single fraction so you have
[cl(N+1)+1]/(N+1)=[cl*N+cl+1]/(N+1)
