Answer:
y= -x+7, b= sqrt(2P/a), c=3P^2-b
Step-by-step explanation:
First, make a table regarding both of the equations. You will eventually find out that both lines intersect at the point (2, 5) after you find the points on the table. From there, subtract x from both sides in the equation x + y = 2. You will get y = -x + 2. Since they said the line was parallel, find a line that has the slope of negative one. Since we know that this line intersects the point in which the first two lines intersect, we know that the y-intercept will be 7. The equation of the line would be y=-x+7.
Multiply both sides by 2. Then, divide both sides by a to get b^2=(2P/a). Take the square root to get the value of b, which is sqrt(2P/a).
Square both sides of the equation to get P^2=(b+c)/3. Cross multiply to get 3P^2=b+c. Subtract b from both sides to get c=3P^2-b.
Answer:
hello attached below is the required table that is missing and the completed table as well
A) mean deviation = 0.1645
B) 2/3
Step-by-step explanation:
From the table we calculated the midpoint value (x) and c.f
N = 27
median class = 2.35 to 2.45
median = l + [ (N/2) - cf / F ] * H
= 2.35 + [ (13.5 - 13) / 6 ] *0.1 = 2.3583
hence mean deviation by median
= summation of fi |xi -M| / summation of fi
= 4.4417 / 27 = 0.1645
B ) probability of getting an odd number or prime number or both
the probability of an odd number = 1/2
also the probability of an even number = 1/2
while the probability of getting neither of them = 1/3
hence the probability of getting odd or prime or both
= 1/2 + 1/2 - 1/3 = 2/3
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>
I can sure try mate!!! What is it that you need help with?
