Answer:
20 +/- $6.74
= ( $13.26, $26.74)
The 90% confidence interval for the difference in average amounts spent on textbooks (math majors - English majors) is ( $13.26, $26.74)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x1-x2 +/- margin of error
x1-x2 +/- z(√(r1^2/n1 + r2^2/n2)
Given that;
Mean x1 = $200
x2 = $180
Standard deviation r1 = $22.50
r2 = $18.30
Number of samples n1 = 60
n2 = 40
Confidence interval = 90%
z(at 90% confidence) = 1.645
Substituting the values we have;
$200-$180 +/-1.645(√(22.5^2/60 +18.3^2/40)
$20 +/- 6.744449847374
$20 +/- $6.74
= ( $13.26, $26.74)
The 90% confidence interval for the difference in average amounts spent on textbooks (math majors - English majors) is ( $13.26, $26.74)
Answer:
The equation is: y = 1/2x + 7
Step-by-step explanation:
The slope of j is -2. The slope of a line perpendicular to line j has a negative inverse: 1/2. The point is (-8, 3).
Use the point slope form of the equation:
y - y1 = m(x - x1)
Substitute:
y - 3 = 1/2(x - (-8))
y - 3 = 1/2x + 4
y = 1/2x + 4 + 3
y = 1/2x + 7
Proof:
Solve for f(x) when x = -8.
f(x) = 1/2x + 7
f(-8) = 1/2(-8) + 7
= -8/2 + 7
= -4 + 7
= 3, giving the point (-8, 3)
notice, the first term is 9.5
from there it goes to 11.5, well, how much is it being added to get 11.5?
well 9.5 + 2, is 11.5, so it was added 2 to 9.5
then it goes from 11.5 to 13.5, namely, 11.5+2 = 13.5
and then 13.5+2 = 15.5 and so on
so, the "common difference" or the common addition value, is 2
how do you find the nth term?
well
Answer: 20
20 since you have to divide 100 by 5 which equals 20 since you have to use the inverse operation of multiplication which is division.
Hope this helps!
