1. The midpoint of the segment joining points (a, b) and ( j, k) is ((j+a)/2,(k+b)/2)
2. Let the coordinate of H be (a, b)
T(0, 4) = ((a + 0)/2, (b + 2)/2)
(a + 0)/2 = 0 => a + 0 = 0 => a = 0
(b + 2)/2 = 4 => b + 2 = (2 x 4) = 8 => b = 8 - 2 = 6
Therefore, the cordinate of H is (0, 6)
3. Point (-4, 3) lies in Quadrant II
4. Point (6, 0) lies on the x-axis
5. Any line with no slope is parallel to the y-axis
7. a is the value of the x-coordinate.
5a + 3 = 8
5a = 8 - 3 = 5
a = 5/5 = 1
a = 1
8. Equation of a circle is given by (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
For the given circle (x + 5)^2 + (y - 7)^2 = 36 => (x - (-5))^2 + (y - 7)^2 = 6^2 => a = -5 and b = 7.
Therefore, its center point is (-5, 7)
9. Equation of a circle is given by (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
For the given circle (x + 5)^2 + (y - 7)^2 = 36 => (x - (-5))^2 + (y - 7)^2 = 6^2 => r = 6.
Therefore, its radius is 6
10. If the equation of a circle is (x - 2)^2 + (y - 6)^2 = 4, the center is point (2, 6).
True
Answer:
0.18 ; 0.1875 ; No
Step-by-step explanation:
Let:
Person making the order = P
Other person = O
Gift wrapping = w
P(p) = 0.7 ; P(O) = 0.3 ; p(w|O) = 0.60 ; P(w|P) = 0.10
What is the probability that a randomly selected order will be a gift wrapped and sent to a person other than the person making the order?
Using the relation :
P(W|O) = P(WnO) / P(O)
P(WnO) = P(W|O) * P(O)
P(WnO) = 0.60 * 0.3 = 0.18
b. What is the probability that a randomly selected order will be gift wrapped?
P(W) = P(W|O) * P(O) + P(W|P) * P(P)
P(W) = (0.60 * 0.3) + (0.1 * 0.7)
P(W) = 0.18 + 0.07
P(W) = 0.1875
c. Is gift wrapping independent of the destination of the gifts? Justify your response statistically
No.
For independent events the occurrence of A does not impact the occurrence if the other.
The area of the surface of the pool is 7322.64 m².
<h3>What is Area of rectangle?</h3>
The area of rectangle is product of length and its breadth.
i.e., length * breadth
Given: Length = 100 m, Width = 52 m and Diameter = 52 m
Area of rectangle,
= 100 × 52
= 5200 m²
Now,
Area of semicircle = 1/2 × πr²
=1/2 × 3.14 × 26²
= 1061.32 m²
Hence, area of the surface of the pool is
= 1061.32 + 1061.32 + 5200
= 7322.64 m²
Learn more about this concept here:
brainly.com/question/15461609
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I thinkkkkk that you have to see what 2 #'s equal 7??
