Answer:
a) For the 90% confidence interval the value of 
and 
, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
b) For the 99% confidence interval the value of 
and 
, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
Step-by-step explanation:
Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
Solution to the problem
Part a
For the 90% confidence interval the value of and , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
Part b
For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
Answer:
-5/2
Step-by-step explanation:
i really hope this helps you out!
The answer is "N". Even though "W" has parallel lines, "N" only has 2.
Hope it helps!
-Lacy
Answer:
r=-11
Step-by-step explanation:
7r+2=5(r-4)
7r+2=5r-20
2r=-22
r=-11
1)
x^2 + 4x - 5
y = --------------------
3x^2 - 12
Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite
=> 3x^2 - 12 = 0 => 3x^2 = 12
=> x^2 = 12 / 3 = 4
=> x = +/-2
Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) = ∞
Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞
Limit of y when x -> - 2(+) = +∞
Limit of y when x -> - 2(-) = -
=> x = 2 and x = - 2 are a vertical asymptotes
Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite
Limi of y when x -> - ∞ = 1/3
Lim of y when x -> +∞ = 1/3
=> Horizontal asymptote y = 1/3
x - intercept => y = 0
=>
x^2 + 4x - 5
0 = ------------------ => x ^2 + 4x - 5 = 0
3x^2 - 12
Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1
=> x-intercepts x = - 5 and x = 1
Domain: all the real values except x = 2 and x = - 2
2) y = - 2 / (x - 4) - 1
using the same criteria you get:
Vertical asymptote: x = 4
Horizontal asymptote: y = - 1
Domain:all the real values except x = 4
Range: all the real values except y = - 1
3)
x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)
factor x^2 - 3x - 10 => (x - 5)(x + 2)
Multiply both sides by (x - 5) (x + 2)
=> x(x - 5) + 7( x + 2) = 14
=> x^2 - 5x + 7x + 14 = 14
=> x^2 + 2x = 0
=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0
Answer: x = 0
