The graph that gives the best representation of the scenario is: option 1 (A).
<h3>Graph of Distance vs Time</h3>
- A typical graph of distance vs time shows the distance covered as against time.
- If the graph shows a line that slopes upwards, it implies movement from a spot, if it shows an horizontal line, it implies a rest.
The first graph is the correct one that best represents the scenario because it starts at the point of origin (0, 0), which means the bird covers not distance at zero time when it is still at its nest.
As the bird flies, distance increases with time till it stopped at a point to eat.. This is represented by horizontal line, meaning time increased but distance from the nest remains the same.
The upward slope from the spot represents the further distance the bird is flying in search for more food.
Therefore, the graph that gives the best representation of the scenario is: option 1 (A).
Learn more about distance vs time graph on:
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<h3>
Answer: -7 < x < 17</h3>
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Explanation:
Plug in the lower bound of the domain, which is x = -3
f(x) = 3x+2
f(-3) = 3(-3)+2
f(-3) = -9+2
f(-3) = -7
If x = -3, then the output is y = -7. Since f(x) is an increasing function (due to the positive slope), we know that y = -7 is the lower bound of the range.
If you plugged in x = 5, you should find that f(5) = 17 making this the upper bound of the range.
The range of f(x) is -7 < y < 17
Recall that the domain and range swap places when going from the original function f(x) to the inverse 
This swap happens because how x and y change places when determining the inverse itself. In other words, you go from y = 3x+2 to x = 3y+2. Solving for y gets us y = (x-2)/3 which is the inverse.
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In short, we found the range of f(x) is -7 < y < 17.
That means the domain of the inverse is -7 < x < 17 since the domain and range swap roles when going from original to inverse.
-- If two lines on a graph are parallel, then their slopes are equal.
-- If two lines on a graph are perpendicular, then their slopes are
negative reciprocals.
The solution for this problem would be:
Given that there is 99.999%.
Let denote n as the network servers and p as the reliability of each server.
So the probability that the network uptime = 1 - (1 - p)^n
Therefore, (1-p) ^n = 0.00001
a. x= log(1-.99999)÷log(1-.97)= 3.2833 is the answer
1-(1-.97)^3= 0.99999 + 0.0001 = 1
b. x = log(1-.99999)÷log(1-.88) = 5.43 is the answer
1-(1-.88)^3= 0.99 + 0.0001 = approx 1
Answer:
Step-by-step explanation:
(x₁, y₁) = (-2 , -5) & (x₂ , y₂) = (-3 , 1)
Midpoint = 
![=(\frac{-2 + [-3]}{2} ,\frac{-5 + 1}{2})\\\\\\=(\frac{-5}{2} , \frac{-4}{2})\\\\=(-2.5, -2)](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B-2%20%2B%20%5B-3%5D%7D%7B2%7D%20%2C%5Cfrac%7B-5%20%2B%201%7D%7B2%7D%29%5C%5C%5C%5C%5C%5C%3D%28%5Cfrac%7B-5%7D%7B2%7D%20%2C%20%5Cfrac%7B-4%7D%7B2%7D%29%5C%5C%5C%5C%3D%28-2.5%2C%20-2%29)
