Answer:
As a fraction, the answer is exactly 86/969
In decimal form, the answer is approximately 0.08875
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Work Shown:
The assumption is that no replacements are made for each selection.
5 red, 6 blue, 8 green
5+6+8 = 19 total
A = P(3 red) = (5/19)*(4/18)*(3/17) = 10/969
B = P(3 blue) = (6/19)*(5/18)*(4/17) = 20/969
C = P(3 green) = (8/19)*(7/18)*(6/17) = 56/969
D = P(3 all same color)
D = A+B+C
D = 10/969 + 20/969 + 56/969
D = (10+20+56)/969
D = 86/969
D = 0.08875
Answer:
234375
hope this helps
have a good day :)
Step-by-step explanation:
Equation 1) y = -2x - 6
Equation 2) y = x + 9
Move equations around, by moving the (x)s to the left side of the equation so that x and y are on the same side.
To do so, add 2x to both sides in equation 1, and subtract x from both sides in equation 2.
1) 2x + y = 6
2) -x + y = 9
Subtract equations from each other.
3x = -3
Divide both sides by 3.
x = -1
Plug in -1 for x in first equation.
y = -2x - 6
y = -2(-1) - 6
Simplify.
y = 2 - 6
y = -4
(-1, -4) ==> x = -1, y = -4
~Hope I helped!~
Answer: x=9 is an extraneous solution
Step-by-step explanation:
Given
Function is 
for function to exist
Therefore, 9 satisfies the above equation. It is it's extraneous solution.
Answer:
B. x < -4 and x > 3
Step-by-step explanation:
Factor and set = to 0
= 0
x = - 4 or x = 3 I call these critical values
The two numbers would divide a number line into 3 intervals. Pick a value in one of the intervals and put it in the original expression. If it makes the function positive, then all the values in that interval make the function positive. If the value you picked makes the function negative, then the values in the other intervals will make the function negative. Let's pick the value of 0 and substitute it into the function
We get 
+ 0 - 12 = -12 which is not positive. Therefore, all the values between -4 and 3 will make the function negative. So, the values less than -4 or greater than 3 will make the function positive. Therefore, B is the correct answer.
Another way to do this problem is to graph the function and see where the graph is above the x-axis. But, sometimes it is not easy to graph the function.
