Answer:
Step-by-step explanation:
Given 
a)
we know that 
therefore 
on integrating we get
c=(1/6640)
b)
on doing the integration we get
=0.37349
c)
marginal density of X is
on doing integration we get
f(x)=(4x+3)/3320 ; 0<x<40
marginal density of Y is
on doing integration we get
d)
solve the above integration we get the answer
e)
solve the above integration we get the answer
f)
Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.
we know f(x,y)
In the (c) bit we got f(x) and f(y)
f(x,y)cramster-equation-2006112927536330036287f(x).f(y)
therefore X and Y are not independent
Answer: the hourly fee for the rototiller is $3
Step-by-step explanation:
Let h represent the hourly fee for renting the rototiller.
The rental company charged an initial fee of $43 with an additional fee per hour. Assuming that they rented the rototiller for x hours. This means that if they rented the rototiller for x hours, the total amount that they would pay is
43 + hx
If they paid $64 after renting the rototiller for 7 hours, the equation that models this problem would be
43 + 7h = 64
7h = 64 - 43 = 21
h = 21/7 = $3
Write in numbers not words
Answer:
diameter of #12 wire = (80.80
0.001) = 0.08080 inches
Step-by-step explanation:
i) #12 wire is 80.80 mils
ii) Therefore diameter of #12 wire = 80.80 mils.
iii) 1 mil = 0.001 inch
iv) therefore diameter of #12 wire = (80.80
0.001) = 0.08080 inches
Answer:
1) ΔCBF ≅ ΔCDF by (SSS)
2) ΔBFA ≅ ΔDFE by (SAS)
3) ΔCBE ≅ ΔCDA by (HL)
Step-by-step explanation:
1) Since BC ≅ DC and DF ≅ BF where CF ≅ CF (reflective property) we have;
ΔCBF ≅ ΔCDF by Side Side Side (SSS) rule of congruency
2) Since DF ≅ BF and FA ≅ FE where ∠DFE = ∠BFA (alternate angles)
Therefore;
ΔBFA ≅ ΔDFE by Side Angle Side (SAS) rule of congruency
3) Since FA ≅ FE and DF ≅ BF then where EB = FE + BF and AD = FA + DF
Where:
EB and AD are the hypotenuse sides of ΔCBE and ΔCDA respectively
We have that;
EB = AD from FE + BF = FA + DF
Where we also have BC ≅ DC
Where:
BC and DC are the legs of ΔCBE and ΔCDA respectively
Then we have the following relation;
ΔCBE ≅ ΔCDA by Hypotenuse Leg (HL).