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seropon [69]
4 years ago
14

What is the answer to this: (2^2)^2/2*2^4

Mathematics
2 answers:
zalisa [80]4 years ago
4 0

Answer:

64

Step-by-step explanation:

sertanlavr [38]4 years ago
3 0

Answer:

64

Step-by-step explanation:

2^2*2^4

4*16

=64

hope it helps

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vredina [299]
No, the sequence is algebraic.
7 0
3 years ago
A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely t
nadezda [96]

Answer:

Expect to lose about $0.73 or 73 cents for each game played

Step-by-step explanation:

Let's define the four events:

F = event of drawing a face card

N = event of drawing a non-face card

H = event of the coin landing on heads

T = event of the coin landing on tails

The events F and N are complementary and this means that one event or the other, but not both, must happen. We either draw a face card (F) or we don't (N). This is why the probabilities add to 1

P(F) + P(N) = 1

Thus;

P(N) = 1 - P(F)

There are 4 suits with 3 face cards per suit (King, Queen, Jack).

So 4 x 3 = 12 face cards out of 52 cards total.

Therefore,

P(F) = probability of drawing a face card = (number of face cards)/(number of cards total)

P(F) = 12/52 = 3/13

And,

P(N) = probability of drawing a non-face card

P(N) = 1 - P(F)

P(N) = 1 - (3/13)

P(N) = 10/13

Now, assuming we have a fair coin with either side is likely to be landed on, it means that;

P(H) = 1/2

P(T) = 1/2

So, P(H) + P(T) = 1

Assuming the events of drawing a card and flipping a coin are independent, then we can form the compound probabilities

P(F & H) = P(F) x P(H)

P(F & H) = (3/13) x (1/2) = 3/26

P(F & T) = P(F) x P(T)

P(F & T) = (3/13)*(1/2) = 3/26

Now, Similar to the probability P(X) notation, let's introduce the function V(X) where V is the net value and X is the general event. To be more specific, writing V(F) represents the net value of drawing a face card.

The three cases we're concerned with are:

V(F & H) = net value for getting face card and heads = 5

V(F & T) = net value for getting face card and tails = 2

V(N) = net value for getting non face card = -2

The negative value (-2) indicates a loss of 2 dollars.

When we play the game out, there are three cases:

Case A = drawing a face card and the coin landing on heads

Case B = drawing a face card and the coin landing on tails

Case C = drawing a non-face card

What we do is multiply the probabilities for each case happening with the net values for each case.

Thus;

For case A, we have the probability P(F & H) = 3/26 and the net value V(F & H) = 5

Hence;

P(F & H) x V(F & H) = (3/26) x 5 = 15/26 = 15/26

Similarly for case B

P(F & T) x V(F & T) = (3/26) x 2 = 6/26 = 3/13

and finally case C

P(N) x V(N) = (10/13) x (-2) = -20/13

Let's now add them up to get;

(15/26) + (3/13) + (-20/13)

This gives; (15 + 6 - 40)/26 = -19/26 = $-0.73

At this expected value, it means that we expect to lose about $0.73 or 73 cents for each game played. This is not a fair game (because expected value isn't 0). Thus, the game clearly favors the house instead of the player.

5 0
3 years ago
5(x + y) -3( x - y)=
exis [7]

5(x + y) - 3(x - y)

5x + 5y - 3x + 3y

5x - 3x + 5y + 3y

2x + 8y

5(x + y) - 3(x - y) = 2x + 8y

7 0
3 years ago
Read 2 more answers
What are the multiples of 5 between 61 and 69​
Blizzard [7]

Answer:

I think it is 65 because we can divide it by 5

8 0
3 years ago
Read 2 more answers
Which of the following has zeros that are not significant?
serious [3.7K]

Answer:

1.30 N

Step-by-step explanation:

All zero's preceding the first integers are never significant. For example: 0.0078 has two significant figures. All zero's after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures

4 0
3 years ago
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