5+2=7 which means there will be 7 zeros so it is 10000000 times greater than
Answer:
The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.7291
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have
. So



The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.
Answer:
Vel_jet_r = (464.645 mph) North + (35.35 mph) East
||Vel_jet_r|| = 465.993 mph
Step-by-step explanation:
We need to decompose the velocity of the wind into a component that can be added (or subtracted from the velocity of the jet)
The velocity of the jet
500 mph North
Velocity of the wind
50 mph SouthEast = 50 cos(45) East + 50 sin (45) South
South = - North
Vel_ wind = 50 cos(45) mph East - 50 sin (45) mph North
Vel _wind = 35.35 mph East - 35.35 mph North
This means that the resulting velocity of the jet is equal to
Vel_jet_r = (500 mph - 35.35 mph) North + 35.35 mph East
Vel_jet_r = (464.645 mph) North + (35.35 mph) East
An the jet has a magnitude velocity of
||Vel_jet_r|| = sqrt ((464.645 mph)^2 + (35.35 mph)^2)
||Vel_jet_r|| = 465.993 mph