Answer:
4.80 mm to 2 d.p
Step-by-step explanation:
w = f(r)
f'(r) = (dw/dr) = (0.0218 mm/mm)
So, for a small difference in rainfall of 220 mm, what is the corresponding small difference in width of leaves in the two forests given.
One definition of a derivative or a rate of change is that it is the ratio of very small differences in the dependent variable to very small differences in the independent variable.
Mathematically,
(dw/dr) = (Δw/Δr) for very small Δw and Δr.
0.0218 = (Δw/220)
Δw = 0.0218 × 220 = 4.796 mm = 4.80 mm to 2 d.p
Hope this Helps!!!
Answer:
Step-by-step explanation:
P(red gum ball)= 6/14=3/7
6/14 can be simplified to 3/7 by dividing by 2. 6/2=3. 14/2=7.
P(red gum ball)=3/7
Answer:
a) 
And we can use the probability mass function and we got:
And adding we got:

b)
c) ![P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]](https://tex.z-dn.net/?f=P%28X%3E3%29%20%3D%201-P%28X%20%5Cleq%203%29%20%3D%201-%20%5BP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%2BP%28X%3D3%29%5D%20)


And replacing we got:
![P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886](https://tex.z-dn.net/?f=%20P%28X%3E3%29%20%3D%201-%5B0.0115%2B0.0576%2B0.1369%2B0.2054%5D%3D%201-0.4114%3D%200.5886)
d) 
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Part a
We want this probability:

And we can use the probability mass function and we got:
And adding we got:

Part b
We want this probability:

And using the probability mass function we got:
Part c
We want this probability:

We can use the complement rule and we got:
![P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]](https://tex.z-dn.net/?f=P%28X%3E3%29%20%3D%201-P%28X%20%5Cleq%203%29%20%3D%201-%20%5BP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%2BP%28X%3D3%29%5D%20)


And replacing we got:
![P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886](https://tex.z-dn.net/?f=%20P%28X%3E3%29%20%3D%201-%5B0.0115%2B0.0576%2B0.1369%2B0.2054%5D%3D%201-0.4114%3D%200.5886)
Part d
The expected value is given by:

And replacing we got:
