Is 400 A whole number a integer a real number or rational number
1 answer:
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The statements that are true about the intervals of the continuous function are Options 2, 4 and 5
- f(x) ≤ 0 over the interval [0, 2].
- f(x) > 0 over the interval (–2, 0).
- f(x) ≥ 0 over the interval [2, ).
Looking at the values given, the intervals which satisfies the condition are known to be:
f(x)<=0 over the interval [0,2]
f(x)>0 over the interval (-2,0)
f(x)>=0 over the interval [2,∞)
Because:
Since the table with x and f(x) values, we have to examine analyze the table and see the each option that is in line with f(x) or not .
Examine the values of x that is from -3 to 3, the f(x) values are both positive and negative . hence f(x)>0 is false over the interval (-∞,3)
Looking at the the interval from 0 to 2, the f(x) values are 0 and negative. Hence, f(x)<=0 over the interval [0,2]
When you look over the interval (-1,1), the f(x) values are said to be both positive and negative and as such, f(x)<0 is false over the interval (-1,1)
When you look at the interval (-2,0) , the f(x) is positive and as such, f(x)>0 over the interval (-2,0)
Looking at the interval [2,∞), f(x) is positive and as such, f(x)>=0 over the interval [2,∞)
Therefore, Option 2, 4 and 5 are correct.
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Answer:
(a) 0.9412
(b) 0.9996 ≈ 1
Step-by-step explanation:
Denote the events a follows:
= a person passes the security system
= a person is a security hazard
Given:
Then,
(a)
Compute the probability that a person passes the security system using the total probability rule as follows:
The total probability rule states that:
The value of P (P) is:
Thus, the probability that a person passes the security system is 0.9412.
(b)
Compute the probability that a person who passes through the system is without any security problems as follows:
Thus, the probability that a person who passes through the system is without any security problems is approximately 1.
Answer:
The vertex of a quadratic equation corresponds to the point where the maximum or minimum value is located.
If the function has a positive leading coefficient, the vertex corresponds to the minimum value.
If it has a negative leading coefficient, the vertex corresponds to the maximum valuevalue
If the vertex is located at
(–2, 0)
The possibilities are
y = (x-2)^2
or,
y = - (x-2)^2
Since the problem tells us the answer, we adopt the positive values
Answer:
y = (x-2)^2
See attached picture
