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:
As exact answers:
and 
As decimal answers:
x = -6.2749172 ≈-6.3
x = 1.27491722 ≈ 1.3
Step-by-step explanation:
For quadratic equations, you can use the quadratic formula. Rearrange the equation to standard from, which is ax² + bx + c = 0.
x² = –5x + 8
x² + 5x - 8 = 0
State the values for "a", "b" and "c",
a = 1; b = 5; c = -8
(Ignore the Â, it's a formatting error)
Simplify
Two negatives make a positive
Split the equation at the ± for plus and minus:
= 1.27491722 ≈ 1.3
= -6.2749172 ≈ -6.3
Therefore the solutions are 1.3 and -6.3.
Answer: i don't know i am sososososososoossoososososososososososososososososoosososos soorry
Step-by-step explanation:
Rudy answered 2 more questions correctly
...........
46x2=92
44x2=88
...........
92-88=4
11+10=?????????????????????
