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Which of the following expressions are equivalent to

Alexus [3.1K]

Answer:

$(2^{6})^{\frac{1}{2} } , 4^{\frac{1}{2} } * 16^{\frac{1}{2} },8^{\frac{1}{2} } * 8^{\frac{1}{2}, \sqrt{16*4}$

Step-by-step explanation:

Some facts to establish before we try to solve: $\sqrt{64} =8$ and another way of writing $\sqrt{64}$ is $64^{\frac{1}{2} }$ . Also order of operations: PEMDAS

For the first option:

We solve what is in the parenthesis: 2*2=4*2=8*2=16*2=32*2=64 so $64^{\frac{1}{2} }$ , which is another way to write $\sqrt{64}$

For the second option: If the square root of 64 is 8, the there is no way we can multiply another value by it.

The third option: When dealing with square roots, we can solve the inside first and then take the square root:

$8^{2} *8^{2} = 64 * 64=$ definitely not 64 (though if you're curious it's 4096)

The fourth: Another way to write the square root is to raise the value to the 1/2 power so

$4^{\frac{1}{2} }* 16^{\frac{1}{2} }= \sqrt{4} *\sqrt{16} = 2*4=8$

Which is the same as $\sqrt{64}$

Fifth: Like the third, we solve the inside of the square root first

16*4= 64 and then $\sqrt{64}$

Sixth: We need a nice whole number (8), and $\sqrt{32}$ has no clean roots, so its result will have decimals. It can't be our answer

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

Todd begins piano practice ar 4:15 p.m. and ends at 5:50 p.m. How long does he practice

kumpel [21]

He practiced for 1 hour and 30 minutes

4 0

4 years ago

Read 2 more answers

2 Points

Over [174]

Answer:

A. Five places to the left

Step-by-step explanation:

Scientific notation is a method used so we can show a very big/small number easier. Rather than putting so many zeroes, we put the number in the format of a number multiplied by 10^x.

Before putting a number into scientific notation, you need to know which number is a significant figures. In this question, there are two significant figure which is 39, all zeroes on the right is insignificant. We can move the decimal to convert the number into 3.9 * 10^5. Basically you need to move the decimal to the left if the exponent positive, but move it to the right if its negative. The exponent is +5 so we move the decimal points 5 places to the left.

6 0

3 years ago

2^5/4=2^5/2^a=2^b=c please help

Rudiy27

You just need to equate the expressions to find the values for a,b and c. First, let's settle that 2^5/4=8, and 8 is 2^3. Also, an important tool here is the indicie rule:
$\frac{ a^{m} }{ a^{n} } = a^{m-n}$
1st expression: 2^5/4=8
2nd expression: It tells us that 2^5/2^a=2^3 (I know that this equals to 8 as all the expressions are equal therefore is the same as our first expression). Using the indicie law, this means that a=2
3rd expression: 2^b must equal to 8 as well, so b=3
4th expression: c=8 as all of the other expressions we have figured out equal to 8
So a=2, b=3 and c=8

6 0

3 years ago