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)
6x-3-9x-2
6x-9x = -3x
-3-2 = -5
the the equation is: -3x-5
4x -(9-3x) = 8x-1
4x -9 + 3x = 8x-1
7x-9 = 8x-1
-9 + 1 = 8x-7x
-8 = x
x = -8
2) would be equal to 19! / 2^9 * 9!
I would think a reflection across the x-axis but I’m not sure