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Help this is due tom

Help this is due tom-example-1
User Jonathan Bravetti
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1 Answer

12 votes
12 votes

Answer:

9. (a) 1/4

(b) remove 3 alphabet cards

(c) add 4 number cards

10. 1/8

Explanation:

Question 9

Given:

  • 8 number cards
  • 10 alphabet cards
  • 6 picture cards

⇒ total number of cards = 8 + 10 + 6 = 24 cards


\mathsf{Probability \ of \ an \ event \ occurring = (Number \ of \ ways \ it \ can \ occur)/(Total \ number \ of \ outcomes)}

(a) P(picture card) = 6/24 = 1/4

(b) P(alphabet card) = 10/24 = 5/12

If we remove 3 alphabet cards from the box

⇒ number of alphabet cards = 10 - 3 = 7

⇒ total number of cards = 24 - 3 = 21

Therefore, P(alphabet card) = 7/21 = 1/3

(c) P(number card) = 8/24 = 1/3

If we add 4 number cards to the box:

⇒ number of number cards = 8 + 4 = 12

⇒ total number of cards = 24 + 4 = 28

Therefore, P(number card) = 12/28 = 3/7

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Question 10

Given:

  • ΔSDR = ΔPRQ
  • Leg length of ΔSDR and ΔPRQ = 4 cm

⇒ area of ΔSDR = area of ΔPRQ = 1/2 × 4 × 4 = 8 cm²

As PR = SD = 4cm

Area of rectangle = (4 + 4) × (12 + 4) = 128 cm²

P(point lies in ΔSDR) = 8/128 = 1/16

P(point lies in ΔPRQ) = 8/128 = 1/16

P(point lies in ΔSDR) OR P(point lies in ΔPRQ) = 1/16 + 1/16 = 2/16 = 1/8

User Oliver Hausler
by
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