Answer:
μ = 0.169
ME = 0.051
Step-by-step explanation:
The confidence interval is:
CI = μ ± ME
So the mean is the middle of the confidence interval, and the margin of error is half the difference.
μ = (0.118 + 0.220) / 2 = 0.169
ME = (0.220 − 0.118) / 2 = 0.051
D
For these type of problems download photo math
It’ll be faster than here with steps!
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>
Answer:
(p ∧ q)’ ≡ p’ ∨ q’
Step-by-step explanation:
First, p and q have just four (4) possibilities, p∧q is true (t) when p and q are both t.
p ∧ q
t t t
t f f
f f t
f f f
next step is getting the opposite
(p∧q)'
<em>f</em>
<em> t</em>
<em> t</em>
<em> t</em>
Then we get p' V q', V is true (t) when the first or the second is true.
p' V q'
f <em>f</em> f
f <em>t</em> t
t <em>t</em> f
t <em>t</em> t
Let's compare them, ≡ is true if the first is equal to the second one.
(p∧q)' ≡ (p' V q')
<em>f f </em>
<em> t t</em>
<em> t t</em>
<em> t t</em>
Both are true, so
(p ∧ q)’ ≡ p’ ∨ q’
