A.) Describe a transformation sequence that will transform quadrilateral ABCD into quadrilateral A’B’C’D’.

Mathematics

1 answer:

Mariulka [41]3 years ago

8 0

a.) The orientation of ABCD is clockwise, as is the orientation of A'B'C'D'. This means the transformation involves a even number of reflections (may be 0). The orientation of AB is North, and the orientation of A'B' is West, so a rotation of 90° CCW (or equivalent) is involved. We can find the point of intersection of the perpendicular bisectors of AA' and BB' (at (-1, -1)) to determine a suitable center of rotation.

ABCD can be transformed to A'B'C'D' by ...

rotation 90° CCW about the point (-1, -1)

b.) Rotation by 90° can also be accomplished by reflection across a diagonal line. Since we want the orientation to remain unchanged, we need another reflection to put the figure into its final position. A suitable alternate sequence for mapping ABCD to A'B'C'D' is ...

reflection across the line y=x
reflection across the line x=-1

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The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is, $\lambda=\frac{1}{\mu}=\frac{1}{7}$ .

The probability density function of <em>X</em> is:

$f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0$

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

$P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx$

$=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148$