To find which measure of variability is greater and which average number of monthly fatalities is higher, you will need to calculate the mean and the mean absolute deviation for both years
The mean will tell us which is generally higher, and the mean absolute deviation will tell us which has a greater variability.
The correct answer is D.
Please see the attached picture for the work.
Hi there!
PART A:
The system of equations we would use would be:
x + y = 22 (amount of items)
3x + 1y = 30 (cost)
Variables:
x = the amount of strawberry wafers bought at the price of $3
y = the amount of chocolate wafers bought at the price of $1
PART B:
To solve, we'll use substitution because we can easily isolate a variable using the first equation.
Work:
x + y = 22 (first equation)
y = 22 - x (isolating a variable)
3x + 1y = 30 (second equation)
3x + (22 - x) = 30 (substituting into the second equation)
2x + 22 = 30 (simplifying)
2x = 8 (subtracting)
x = 4 strawberry wafers
4 + y = 22 (substituting x into the first equation to solve for y)
y = 18 chocolate wafers
ANSWER:
They bought 4 strawberry wafers and 18 chocolate wafers.
Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
Answer:
160
Step-by-step explanation:
This is a typo and vaguely phrased. However, the closest to a direct solution is to think of it as follows: 35% of the cars are speeding. Out of these, 52% of the cars are also speeding. Hence, the result of cars that are both speeding and speed cars is 0.52*0.35=0.182
Answer:
0.0266, 0.9997,0.7856
Step-by-step explanation:
Given that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11. Let x be the score by A students and Y the score of B.
A)
B) Since X and Y are independent we have
X-Y is Normal with mean = 140-120 =20 and 

C) For a group of 3, average has std deviation = 
