Answer:
1) f(x)<x+4
f(x)>-x-3
f(x)<5
![ANSWER: (-3,4)](https://tex.z-dn.net/?f=ANSWER%3A%20%28-3%2C4%29)
2)2.5>0+2
2.5<0+3
2.5<0+5
![ANSWER: (0,2.5)](https://tex.z-dn.net/?f=ANSWER%3A%20%280%2C2.5%29)
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HOPE IT HELPS
HAVE A GREAT DAY!!
Answer:
2
Step-by-step explanation:
Given
See attachment for chart
Required
Number of off days
To do this, we simply calculate the expected value of the chart.
This is calculated as:
![E(x) = \sum x * f](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Csum%20x%20%2A%20f)
Where
x = days
f = chances
So, we have:
![E(x) = 0* 0.3 + 1 *0.2 + 2 * 0.25 + 3 * 0.2 + 4 * 0.02 + 5 * 0.01 + 6 * 0.01 + 7 * 0.01](https://tex.z-dn.net/?f=E%28x%29%20%3D%200%2A%200.3%20%2B%201%20%2A0.2%20%2B%202%20%2A%200.25%20%2B%203%20%20%2A%200.2%20%2B%204%20%20%2A%200.02%20%2B%205%20%2A%200.01%20%2B%206%20%2A%200.01%20%2B%207%20%2A%200.01)
![E(x) = 1.56](https://tex.z-dn.net/?f=E%28x%29%20%3D%201.56)
![E(x) \approx 2](https://tex.z-dn.net/?f=E%28x%29%20%5Capprox%202)
Range is 6 and the mode is 8
For this case, the main function is given by:
We can apply the following transformation:
Horizontal displacements:
Assume h> 0:
To graph f (x + h) move the graph of f (x) h units to the left.
For h = 1 we have:
Answer:
the graph y=6(x+1)^2 is the graph y=6x^2 moved 1 unit to the left.