Answer:

Step-by-step explanation:
Given <u>v</u> = < - 5, 9 > then the magnitude is
| <u>v</u> | =
=
= 
Answer:
The probability that the plane is oveloaded is P=0.9983.
The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.
Step-by-step explanation:
The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.
If the plane is full, we have 41 men in the plane. This is our sample size.
The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.
So the mean of the sample is 180.5 lb (equal to the population mean).
The standard deviation is:

Then, we can calculate the z value for x=163 lb.

The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017

Then the probability that the plane is oveloaded is P=0.9983:

The pilot should take out the baggage or have less passengers in the plain to not overload.
Answer: The z score that corresponds to the value x=22 is 0.4 .
Step-by-step explanation:
Given : A normal distribution has a mean 20 and standard deviation 5.
i.e. 

Let x be the random selected variable.
We know that to find the z-score corresponds to the value x is given by :-

For x = 22, we have

Hence, the z score that corresponds to the value x=22 is 0.4
Answer: <span>A. about 30.8 miles
</span><span>
The x and y distance from the campsite at A (3, – 5), and the next checkpoint station is located at B (–10, 1) would be:
Xab= Xb- Xa= -10 - 3= -13
Yab= Yb- Ya= 1- (-5)= 6
The resultant distance would be
c^2= a^2+b^2
c^2= (-13)^2 + (6)^2
c= </span>√(169+36)= √205
c=14.32
If each unit in the coordinate plane represents 2.15 miles, the true distance= 14.32*2.15 miles= 30.78 miles
I’m not 100% sure so sorry if it’s wrong but I think it’s 3/2