If we calculate the net take home pay and we assume the employer withheld federal income tax (wage-bracket, married, 2 <span>allowances), social security taxes, and state income tax (2%)
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Married at least $500 but not more than $510 $21
Social Security at 4.2% $21
State income tax at 2% $10
Total taxes: $52
Total net take-home pay: $598
Answer:
Im doing it right now one second
Step-by-step explanation:
Answer:
Step-by-step explanation:
Researchers measured the data speeds for a particular smartphone carrier at 50 airports.
The highest speed measured was 76.6 Mbps.
n= 50
X[bar]= 17.95
S= 23.39
a. What is the difference between the carrier's highest data speed and the mean of all 50 data speeds?
If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps
b. How many standard deviations is that [the difference found in part (a)]?
To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation
Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations
c. Convert the carrier's highest data speed to a z score.
The value is X= 76.6
Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51
d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.
I hope it helps!
Answer:
D
Step-by-step explanation:
Using the cosine ratio in the right triangle.
cos66° = 
= 
( multiply both sides by x )
x × cos66° = 13 ( divide both sides by cos66° )
x = 
≈ 32.0 ( to the nearest tenth )
I believe you are correct
