Answer:
d =7.6
Step-by-step explanation:

Answer: 44
Explanation: it is divisible by two but does not end with two
Answer:
The answer is 3.64m
This is because if you draw a diagram, and label all the sides of the triangle (opp, hyp, adj), the adjacent angle is 3m. You can now use the sine rule to find the hypotenuse (length of ladder) by doing: cos 30= 3/h. You divide cos 30 by 3 and you get the answer of 3.64m rounding to 2 decimal places.
Sorry i don’t know if it’s me but i can’t see the picture.
Answer:
-764.28
Step-by-step explanation:
Given the joint cumulative distribution of X and Y as

#First find
and probability distribution function ,
:

#Have determined the probability distribution unction ,
, we calculate the Expectation of the random variable X:

#We then calculate
:

Hence, the Var(X) is 764.28