Answer:
range=u ± 3.09 sd
Step-by-step explanation:
Given:
mean, u= 26.8 mpg
standard deviation, sd=72 mpg
% contained in interval = 99.8%
the interval for 99.8% of the values of a normal distribution is given by
mean ± 3.09 standard deviation= u ± 3.09 sd
=26.8 ± 3.09(72)
=26.8 ± 222.48
= 249.28 , -195.68
range=u ± 3.09 sd = 249.28 , -195.68 !
The answer is shown below:
Rectangle JKLM has an area of 36 square centimeters. The side is at least 4 centimeters long. Prove: KL is 9 centimeters. Assume that a. "KL > 0". Then the area of rectangle JKLM is greater than b.) "40 centimeters² ", which contradicts the given information that c. the area is 40 centimeters². So the assumption must be false. Therefore, d.) "KL <= 10".
Below are I assume the choices, it can be found elsewhere:
A. 15x+10y≥250
B. x+y≤250
C. x+y≥20
D. 15x+10y≥20
E. x+y≤20
<span>F. 15x+10y≤250
</span>
The inequalities that represent constraints for this situation is 15x+10y<=250 and <span>x+y>20. Thank you for posting your question here. I hope the answer helps. </span>
If we draw the contingency table of x (vertical) against y (horiz.), we have a square.
For n=4, we have (legend: < : x<y = : x=y > : x>y
y 1 2 3 4
x
1 = < < <
2 > = < <
3 > > = <
4 > > > =
We see that there are n(n-1)/2 cases of x<y out of n^2.
Therefore,
p(x<y)=n(n-1)/(2n^2)=(n-1)/(2n)
However, if the sample space is continuous, it will be simply p(x<y)=1/2.
Answer:
483
Step-by-step explanation:
840 minus 357 is 483
