I had the same thing to do :) im doing mine now, if needed just pm me and ill help more, just right about the proposes you see
Answer:
Refer below
Step-by-step explanation:
a) a and b are the lower and higher values of the interval for which uniform distribution is defined.
Here a= 6 and b =10
b) Mean of the uniform distribution= (a+b)/2 = (6+10)/2 =8
Or int x (1/4) dx = x^2/8 = 8
c) Variance of the uniform distribution = (b^2-a^2)/12 = (100-64)/12
= 36/12 =3
Std dev = sq rt of 3 = 1.732
d) To find total area
PDF of the distribution = 1/(b-a) = 1/4, 6<x<10
Area = \int 6 to 10 of 1/4 dx
= x/4
Subtitute limits
= (10-6)/4 =1
So total area = 1
d)P(X>7) = int 7 to 10 of 1/4 dx = 3/4
e) P(7<x<9) = Int 7 to 9 of 1/4 dx = 2/4 = 1/2
Answer:
B) linear function with a positive rate of change
Step-by-step explanation:
Each increase by 1 m^2 in area produces the same 0.2 kW increase in power, so the relationship between area and power is linear. Since they both go up (or both go down), the "rate of change" is positive. (If it were negative, one would go down when the other went up.)
The best description is that of B.
William needs 102.6 fencing in all because 76.9 plus 25.7 equals 102.6, hope this helps!
