A : for x = 4 y = 16 not 6 FALSE
B : for x = 4 y = 20 not 6 FALSE
C : for x = 4 y = 11 not 6 FALSE
D : for x = 4 y = x + 2 = 6 is
6 The Last one verifying the first row of the table
pls mark as brainliest
Order of operation
PEMDAS-Parentheses,Exponents,Multiplication,Division,Addition,Subtraction
So 7*2=14
Then 36-14=22
22 is your answer and that's how you solve it :)
Answer: 6x + 126
Hope this helped.
Answer:
0.3413 = 34.13% probability that the crew member earns between $20.50 and $24.00 per hour
Step-by-step explanation:
Problems of normal distributions 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.
A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was $20.50, with a standard deviation of $3.50.
This means that 
a. Between $20.50 and $24.00 per hour
This is the pvalue of Z when X = 24 subtracted by the pvalue of Z when X = 20.5. So
X = 24



has a pvalue of 0.8413
X = 20.5



has a pvalue of 0.5
0.8413 - 0.5 = 0.3413
0.3413 = 34.13% probability that the crew member earns between $20.50 and $24.00 per hour
Answer:
The null hypothesis would be
H0: μ1=μ2=μ3 (all the means of the three sites are equal)
Step-by-step explanation:
Site I Site II Site III
28 47 46
54 22 27
44 65 40
16 56 22
21 17 58
36
The null hypothesis would be
H0: μ1=μ2=μ3 (all the means of the three sites are equal) against the claim
Ha: μ1≠μ2≠μ3 (all the means of the three sites are not equal)
Which if H0 is true, has an F -distribution with v1=k-1 and v2= n-k degrees of freedom.
ANOVA (one way of analysis of variance ) can be applied to test this hypothesis.