En una tienda se anuncia la rebaja de 20% en todos sus artículos . Rafael elije una camisa y al llegar a la caja le informan que
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Answer:
Please find attached the required plot accomplished with an online tool
Part A:
1/4
Part B:
P''(-1, 0), Q''(0, -1), and R''(2, -1)
Part C:
Triangle PQR is similar to triangle P''Q''R'' but they are not congruent
Step-by-step explanation:
Part A:
Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)
Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)
The dimensions of the sides of the triangle are given by the relation;
Where;
(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment
For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);
By substitution into the length equation, we get;
The length of segment PQ = 4·√2
The length of segment PR = 4·√10
The length of segment RQ = 8
The length of segment P'Q' = √2
The length of segment P'R' = √10
The length of segment R'Q' = 2
Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4
Part B:
Reflection of (x, y) across the y-axis gives;
(x, y) image after reflection across the y-axis = (-x, y)
The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;
P'(1, 0) image after reflection across the y-axis = P''(-1, 0)
Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)
R'(-2, -1) image after reflection across the y-axis = R''(2, -1)
Part C:
Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.
Hello!
For this uniform distribution, the shape of the density function is a line. We can begin by identifying key elements of the distribution:
Range: 6 to 14 pounds
Formula of density function:
y = 1/Range
1 - (14 - 6) = 1/8 = 0.125
f(x) = 0.125
1)
P(x = 7)
A uniform distribution is a continuous distribution, so the probability of x being an EXACT value is approximately 0.
<u>This is the same as if we took the following integral. (Finding the area under a curve with the same start and end point)</u>
Therefore, P(x = 7) = 0.
2)
We can think of this as finding the area of a rectangle.
The width would be the difference between 8.25 pounds and 12 pounds:
W = 12 - 8.25 = 3.75
The height would be the function (y = .125), or:
h = .125
Using the equation for the area of a rectangle: (A = h * w)
A = 3.75 * .125 = 0.469
Thus, P(8.25 < x < 12) = 0.469
3)
To find the probability GREATER than 10.50 pounds, we can subtract this value from the max value for the width:
W = 14 - 10.50 = 3.50
The height is the same as above:
h = .125
Solve for the area:
A = 3.50 * .125 = 0.4375
P(x > 10.5) = 0.4375