Esmerelda flips a coin and draws a card. What is the possibility that she gets “heads” and a black card?

2 answers:

7 0

The possibility of flipping heads is 1 in 2 (1/2)
the possibility of pulling a black card is 26 in 52 (26/52) which equals 1 in 2 (1/2)
multiply those together and you get 1 in 4, 1/4, or a 25%

nalin [4]2 years ago

3 0

Answer:

1/4

Step-by-step explanation:

The probability of flipping a coin and landing a heads is 1/2. Their are 26 black cards in a standard deck of 52 cards- 26/52 (1/2). Multiply the fractions to get the probability.

1/2 * 1/2 = 1/4

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Usimov [2.4K]

Answer:

The correct option is;

C. 52.2 × 4.4 × 2 + 47.8 × 4.4 × 2

Step-by-step explanation:

The area of the square frame is the area of the inner square subtracted from the area of the outer square

Which gives;

52.2² - 47.8² = 440

Therefore;

Option A. 52.2² - 47.8² is a correct way of finding the area of the square frame

Option B. 2 × ((52.2×2.2) + (47.8×2.2)) = 440, is a correct way of finding the area of the square frame

Option C. 52.2 × 4.4 × 2 + 47.8 × 4.4 × 2 = 880, is not a correct way of finding the area of the square frame

The error is the thickness is 2.2 not 4.4

Option D. 100×(52.2 - 47.8) = (52.2 + 47.8) × (52.2 - 47.8) = 52.2² - 47.8² = 440, is a correct way of finding the area of the square frame

5 0

2 years ago

A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate

neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

$C-2=\frac{1}{11}*(C+D-2)...(1)$

We are also told if he instead added 4 glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

$C=\frac{1}{7}*(C+D+4)...(2)$