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3 years ago

Describe the composite transformation that has occurred.

Mathematics

1 answer:

Sidana [21]3 years ago

6 0

Answer:

rotate CCW 90°, reflect across the x-axis
(x, y) ⇒ (-y, -x) . . . . . both transformations together

Step-by-step explanation:

The vertex order ABC is clockwise in the original figure and also in the first image: A'B'C'. The altitude from AC to B is up in the original and left in the first image, indicating a rotation 90° CCW.

The first transformation is a rotation 90° CCW.

The vertex order of A''B''C'' is CCW, indicating a reflection. The direction of the altitude from A''C'' to B'' is still to the left, so the reflection must be over a horizontal line. We find the x-axis bisects the segments A'A'', B'B'', and C'C'', confirming that it is the line of reflection.

The second transformation is reflection across the x-axis.

_____

Algebraically, the transformations are ...

1st: (x, y) ⇒ (-y, x)

2nd: (x, y) ⇒ (x, -y)

Both together: (x, y) ⇒ (-y, -x).

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Answer:

a) $P(X=13)=(24C13)(0.4)^{13} (1-0.4)^{24-13}=0.0608$

$P(X=17)=(24C17)(0.4)^{17} (1-0.4)^{24-17}=0.0017$

$P(X=13 U X=17) = 0.0608+0.0017=0.0624$

b) $P(X \geq 17)=0.000427$

c) $P(X \geq 16)= 1-0.000427=0.9996$

d) $E(X)= np = 22*0.4=8.8$

e) $Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=24, p=0.4)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

3) Part a

$P(X=13)=(22C13)(0.4)^{13} (1-0.4)^{22-13}=0.0336$

$P(X=17)=(22C17)(0.4)^{17} (1-0.4)^{22-17}=0.00035$

$P(X=13 U X=17) = 0.0336+0.00035=0.0340$

4) Part b

$P(X \geq 17)=P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)$

$P(X=17)=(22C17)(0.4)^{17} (1-0.4)^{22-17}=0.00035$

$P(X=18)=(22C18)(0.4)^{18} (1-0.4)^{22-18}=0.0000651$

$P(X=19)=(22C19)(0.4)^{19} (1-0.4)^{22-19}=0.00000914$

$P(X=20)=(22C20)(0.4)^{20} (1-0.4)^{22-20}=0.000000914$

$P(X=21)=(22C21)(0.4)^{21} (1-0.4)^{22-21}=5.81x10^{-8}$

$P(X=22)=(22C22)(0.4)^{22} (1-0.4)^{22-22}=1.76x10^{-9}$

$P(X \geq 17)=0.000427$

5) Part c

$P(X \leq 16)=1-P(X>16)=1-P(X\geq 17) = 1-[P(X=17)+P(X=18)+P(X=19)+P(X=20)+P(X=21)+P(X=22)]$

$P(X \geq 16)= 1-0.000427=0.9996$

6) Part d

The expected value is:

$E(X)= np = 22*0.4=8.8$

7) Part e

The variance is given by:

$Var(X) = np(1-p)= 22*0.4*(1-0,4)=5.28$

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Answer:

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Step-by-step explanation:

I hope I helped you! :)

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