Answer:
$1226.78<x<$1301.22
Step-by-step explanation:
The formula for calculating confidence interval is expressed as;
CI = xbar ± z×(s/√n)
Given
Mean (xbar) = $1264
z is the z score at 90% CI = 1.645
s is the standard deviation = $150
n is the sample size = 44
Substitute
CI = 1264±1.645(150/√44)
CI = 1264±1.645(150/6.63)
CI = 1264±1.645(22.624)
CI = 1264±37.22
CI = (1264-37.22, 1264+37.22)
CI = (1226.78, 1301.22)
Hence the confidence interval of the mean is $1226.78<x<$1301.22
Answer:
x = - 1
Step-by-step explanation:
(8x + 12) = 3 ( multiply both sides by 4 to clear the fraction )
3(8x + 12) = 12 ( divide both sides by 3 )
8x + 12 = 4 ( subtract 12 from both sides )
8x = - 8 ( divide both sides by 8 )
x = - 1
Answer:
K=(3,6)
Step-by-step explanation:
If you reflect it over the y-axis the first number will change to a positive and the second number will stay the same.
Hope this helps have a great day! :)
Answer:
9
Step-by-step explanation:
f(3) = 2(3) + 3
= 9
Answer:
7.64% probability that they spend less than $160 on back-to-college electronics
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

Probability that they spend less than $160 on back-to-college electronics
This is the pvalue of Z when X = 160. So



has a pvalue of 0.0763
7.64% probability that they spend less than $160 on back-to-college electronics