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asambeis
2 months ago
8

Which shows the image of ΔRST after the rotation (x, y) → (y, –x)? On a coordinate plane, a triangle has points R (negative 2, 1

), S (1, 3), T (negative 1, 8). On a coordinate plane, a triangle has points R prime (1, 2), S prime (3, negative 1), T prime (7, 1). On a coordinate plane, a triangle has points R prime (1, negative 2), S prime (3, 1), T prime (7, negative 1). On a coordinate plane, a triangle has points R prime (negative 1, negative 2), S prime (negative 3, 1), T prime (negative 7, negative 1). On a coordinate plane, a triangle has points R prime (2, 1), S prime (negative 1, 3), T prime (1, 7).
Mathematics
2 answers:
PIT_PIT [12.4K]2 months ago
4 0

Answer:

On a coordinate grid, a triangle is defined by the points R' (1, 2), S' (3, -1), T' (7, 1)

Step-by-step explanation:

We can interpret the coordinates of the triangle's vertices as...

R(-2, 1), S(1, 3), T(-1, 7)

Applying the transformation (x, y) ⇒ (y, -x), these coordinates change to...

R'(1, 2), S'(3, -1), T'(7, 1) . . . . . (this corresponds to the first option)

lawyer [12.5K]2 months ago
3 0

Answer:

it's the inverted triangle located on the x-axis

Step-by-step explanation:

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Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.
AnnZ [12381]

A="b is situated in the center"

B="c lies to the right of b"

C="The letters def occur sequentially in that arrangement"

a) b can occupy 7 positions; however, only one of these is the center. Therefore, P(A)=1/7

b) Let X=i; "b holds the i-th position"

Y=j; "c occupies the j-th position"

P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}

P(B)=1/2

c) Let X=i; "d holds the i-th position"

Y=j; "e occupies the j-th position"

Let Z=k; "f is in the i-th position"

P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}

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P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

Events A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

Events A and C are not independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

Events B and C are not independent

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Answer:

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