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sveta [45]
1 year ago
5

Figure ABCD is transformed to obtain figure A′B′C′D′:

Mathematics
1 answer:
xz_007 [3.2K]1 year ago
3 0

Answer:

<u>Part A</u>

  1. Reflect over the y-axis:  (x, y) → (-x, y)
    A (-4, 4) → (4, 4)
    B (-2, 2) → (2, 2)
    C (-2, -1) → (2, -1)
    D (-4, 1) → (4, 1)
  2. Shift 4 units down:  (x, y-4)
    (4, 4-4) → A' (4, 0)
    (2, 2-4) → B' (2, -2)
    (2, -1-4) → C' (2, -5)
    (4, 1-4) → D' (4, -3)

<u>Part B</u>

Two figures are <u>congruent</u> if they have the same shape and size.  (They are allowed to be rotated, reflected and translated, but not resized).

Therefore, ABCD and A'B'C'D' are congruent.  They are the same shape and size as they have only be reflected and translated.

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Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

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n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

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⇒ n(B) = 4,

Since,

\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

(a) So, the probability of B,

P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

\implies P(A\cap B\cap C)=\frac{1}{8}

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