Part A;
There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by <span>(-3, -4), (-4, 3), (1, -2) and (5, -4) can serve.
An example of such system of equation is
x > 0
y > 0
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the x-axis and to the right of the y-axis is shaded.
Part 2:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(2, 2) into the system we have:
2 > 0
2 > 0
as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.
Part C:
Given that </span><span>Natalie
can only attend a school in her designated zone and that Natalie's zone is
defined by y < −2x + 2.
To identify the schools that
Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.
For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true
For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true
For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false
For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true
For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false
For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D.
</span>
34 is larger than 3, so 34.653
Answer:
0.156
Step-by-step explanation:
Using binomial probability formula, we have :
P( a out of n ) =ⁿCₐ x pᵃ x qⁿ⁻ᵃ ------------------------------------------------- (1)
Where n = total number of sample
a = number of success
p = probability of success
q = probability of failure
n-a = number of failures
From the question:
n =10 , a = 7, p=0.54, q = 1-p = 0.46
Substituting into equation (1) we have:
P (7 out of 10) = ¹⁰C₇ x (0.54)⁷x (0.46)¹⁰⁻⁷
= 0.1563
≈ 0.156
90 1 m/s2 189087536
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None of the choices are equivalent
64/4 is 16
a is 4
b is 4
c is 2
d is 1
