Graph the six terms of a finite series where a1 = -3 and r = 1.5.
2 answers:
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This will be a 4th degree polynomial. Our root of x = 7 in factorization form is (x-7). Our root of x = -11 in factorization form is (x+11) and the last one is a complex number. According to the conjugate root theorem, if we have 2+8i, we also HAVE to have 2-8i. In factorization form that first one is (x-(2+8i)) which simplifies to (x-2-8i). Its conjugate in factorization form is (x-2+8i). Now we will FOIL all that out. Let's start with the (x-2-8i)(x-2+8i). That multiplies out to 
. We have to combine like terms here to shorten that a bit. 
. i^2 is equal to -1, and -1(64) = -64. Now we have 
. That is 
. Now let's FOIL in another factorization. 
. That comes out to 
. One more term to go! 
. That, finally, is 
.
Answer:
a. The 95% confidence interval for the difference in mean is C.I. = -3.6 < μ₂ - μ₁ < 4.96
B. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval
c. The assumptions made are;
The variance of the two distributions are equal
Step-by-step explanation:
The given parameters are;
The mean for the 11 students with an eating disorder, = 13.82
The standard deviation for the 11 students with an eating disorder, s₁ = 4.92
The mean for the 14 students who do not have an eating disorder, = 13.14
The standard deviation for the 14 students with an eating disorder, s₂ = 5.29
a. The 95% confidence interval for the difference in mean is given as follows;
The pooled standard deviation, is therefore;
Therefore;
Where at degrees of freedom, df = n₁ + n₂ - 2 = 25 - 2 = 23 the critical-t = 2.07
\left (13.82- 13.14 \right )\pm 2.07 \times\sqrt{5.13241^2 *(\dfrac{1}/{11}+\dfrac{1}{14}\right)}
Therefore, we get;
C.I. = -3.6 < μ₂ - μ₁ < 4.96
b. Therefore, given that the confidence interval extends from positive to negative, therefore, there is a possibility that there is no difference between the mean FNE scores for bulimic and normal students
B. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval
c. The assumptions made are;
The variance of the two distributions are equal