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1 year ago

A quotient is considered rationalized if its denominator contains no

Mathematics

1 answer:

FinnZ [79.3K]1 year ago

4 0

Answer:

radicals or complex numbers

Step-by-step explanation:

A "rationalized" denominator consists entirely of a real number or variable expression with real coefficients. Preferably, any number(s) will be integer(s).

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In the diagram above (not to scale), Vx/y = 12.57 meters/second and

EastWind [94]

The first vector's speed is 12.57 meters for one second
The second vector's speed is 5.21 meters for one second
The total distance is 12.57+5.21 = 17.78 meters over two seconds
Hence the speed is 17.78 / 2 = 8.89m/s

8 0

2 years ago

Read 2 more answers

Carbon–14 is a radioactive isotope that decays exponentially at a rate of 0.0124 percent a year. How many years will it take for

stepan [7]

The decay rate should have units, it should be negative and it should be 100 times smaller than what you posted.
k = -.000124 / years

k = -.000124 / years
Half-Life = ln (.5) / k
Half-Life = -.693147 / -.000124
Half-Life = 5589.8951612903 
Half-Life = 5,590 (rounded)

elapsed time = half-life * log(bgng amt / end amt) / log(2)
elapsed time = 5,590 * log(10) / 0.3010299957 
elapsed time = 5,590 * 1 / 0.3010299957 
elapsed time = 18,569 years

Source:
http://www.1728.org/halflife.htm

5 0

2 years ago

Read 2 more answers

In the equation below, b and c represent the price per pound, in dollars, of beef and chicken, respectively, x weeks after july

Bumek [7]

Answer:

Step-by-step explanation:

Since 1 and c are supposed to be equal, I put 2.35+0.25x=1.75+0.4x and solved that for the x. I got 4. Then, since I want to find the price of the beef, I plugged the 4 in the equation of "b" and got 3.35.

I made many mistakes when solving so I restarted many times which is why it took so long to answer. Whew, I actually found the answer though!

If you want to help, mark me brainliest, I'm only 1 away from getting virtuoso! That would make my day, and hope I made yours.

Bye!

4 0

2 years ago

In the envelope game, there are two players and two envelopes. One of the envelopes is marked ''player 1 " and the other is mark

svetoff [14.1K]

Answer:

Step-by-step explanation:

a) The game tree for k = 5 has been drawn in the uploaded picture below where C stands for continuing and S stands for stopping:

b) Say we were to use backward induction we can clearly observe that stopping is optimal decision for each player in every round. Starting from last round, if player 1 stops he gets $3 otherwise zero if continues. Hence strategy S is optimal there.

Given this, player 2’s payoff to C is $3, while stopping yields $4, so second player will also chooses to stop. To which, player 1’s payoff in k = 3 from C is $1 and her payoff from S is $2, so she stops.

Given that, player 2 would stop in k = 2, which means that player 1 would stop also in k = 1.

The sub game perfect equilibrium is therefore the profile of strategies where both players always stop: (S, S, S) for player 1, and (S, S) for player 2.

c) Irrespective of whether both players would be better off if they could play the game for several rounds, neither can credibly commit to not stopping when given a chance, and so they both end up with small payoffs.

i hope this helps, cheers

5 0

1 year ago

Player V and Player M have competed against each other many times. Historical data show that each player is equally likely to wi

Helen [10]

Answer:

0.46 (46%)

Step-by-step explanation:

We have the following data:

- Probability that player V wins the first set:

$p(1)=0.50$

(because the text says the two players are equally likely to win the first set)

- Probability that player V wins the 2nd set if he has won the 1st set:

$p(2|1)=0.60$

So, the probability that player V wins the first 2 sets is:

$p(12)=p(1)\cdot p(2|1)=(0.50)(0.60)=0.30$ (1)

Instead, the probability that player V loses the 2nd set if he has won the 1st set is 0.40 (=1-0.60), so

$p(2^c|1)=0.40$

So, the probabiity that player V winse the 1st set but loses the 2nd set is

$p(12^c)=p(1)\cdot p(2^c|1)=(0.50)(0.40)=0.20$ (2)

Also, we have:

- Probability that player V loses the 1st set:

$p(1^c)=0.50$

- Probability that she will lose the 2nd set in this case is 0.70, it means that the probability that she will win the 2nd set if she lost the 1st set is 0.30, so:

$p(2|1^c)=0.30$