Answer:
(a)

(b)


Step-by-step explanation:
Given

Solving (a): The pmf
This means that we list out the probability of each value of x.
To do this, we simply subtract the current probability value from the next.
So, we have:

The calculation is as follows:





The x values are gotten by considering where the equality sign is in each range.
means 
means 
means 
means 
means 
Solving (b):

This is calculated as:

From the given function


So:



This is calculated as:




Answer:
An instructor at an army base wants to know if a new program that teaches gun safety is effective is described below in brief details.
Step-by-step explanation:
Although of your shooting practice, the handgun safety program illustrates the center fundamentals of handgun protection so that you can tour the range with courage and with safety in the brain. Study how to accurately control, transport, and deposit your handgun, as well as how to keep children secure around handguns.
The value of the derivative of functions h'(6) as requested in the task content is; 55.
<h3>What is the value of h'(6)?</h3>
Since it follows from the task content that the function h(x)=4f(x)+5g(x)+1.
Hence, the derivative of h(x) can be evaluated as;
h'(x)=4f'(x)+5g'(x)
On this note, by substitution, it follows that;
h'(6)=4(5)+5(7)
h'(6) = 55.
Read more on functions;
brainly.com/question/6561461
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Answer:
1
/27
1 over 27
Step-by-step explanation:
brainliest please?