Answer:

See explanation below.
Step-by-step explanation:
For this case we define first some notation:
A= A new training program will increase customer satisfaction ratings
B= The training program can be kept within the original budget allocation
And for these two events we have defined the following probabilities

We are assuming that the two events are independent so then we have the following propert:

And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:

And using the De Morgan laws we know that:

So then we can write the probability like this:

And using the complement rule we can do this:

Since A and B are independent we have:

And then our final answer would be:


Actually Welcome to the concept of Parallel lines.
We must first understand that, Parallel Lines always have a same Slope, hence the 'm' value in y=mx+c equation will same, here it is '1/2' in the above equation,
so the points here are (-6,-17)
==>
(y-(-17)) = m(x-(-6))
==>
here m = 1/2 ,hence
y+17 = 1/2(x+6)
==> y+17 = 1/2(x) + 3
==> y = 1/2(x) + 3 - 17
==> y = 1/2(x) - 14
hence the Option 4.) is the correct answer!!
Most of the graphing resources used shows that the graph is translated up by 3.25 (3 + 1/4). If you add the 3 and the 1/4, you get the equation of a line:
f(x) = x + 3.25,
which means that the graph of f(x)= x now has a y intercept of 3.25.
The graphs are below. The red line represents f(x) = x, and the green line represents (x+3)+1/4. Hope this helps!
The answer is <span>2(–4y + 13) – 3y = –29
Step 1: Express </span><span>x from the second equation
Step 2: Substitute x into the first equation:
The system of equations is:
</span><span>2x – 3y = –29
x + 4y = 13
Step 1:
</span>The second equation is: x + 4y = 13
Rearrange it to get x: x = - 4y + 13
Step 2:
The first equation is: 2x – 3y = –29
The second equation is: x = - 4y + 13
Substitute x from the second equation into the first one:
2(-4y + 13) - 3y = -29
Therefore, the second choice is correct.