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SashulF [63]
2 years ago
9

What is (3²-2³)⁶⁶ total​

Mathematics
1 answer:
horrorfan [7]2 years ago
6 0

Answer:

1

Step-by-step explanation:

(3²-2³)⁶⁶ = (9 - 8)⁶⁶ = (1)⁶⁶ = 1

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Which is the graph of the linear inequality y < 3x + 1? On a coordinate plane, a solid straight line has a positive slope and
stepan [7]

For this case we have the following inequality: y < 3x + 1 < br/ >

What we must do is to evaluate a point of the Cartesian plane and verify if it is in the shaded region.

The shaded region represents the solution of the system of equations.

For the point (0, 0) we have:

0 < 3(0) + 1 < br / >

0 < 0 + 1 < br / >

0 < 1 < br / >

Therefore, the point (0, 0) is in the shaded region because it satisfies the inequality.

Then, the points that are on the line, are not part of the solution because the sign is of less strict.

Hope I helped ~~Laurel

6 0
2 years ago
Read 2 more answers
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
2 years ago
What is the true solution to l n 20 + l n 5 = 2 l n x x = 5 x = 10 x = 50 x = 100
Pani-rosa [81]

Answer:

x = 10

Step-by-step explanation:

l n 20 + l n 5 = 2 l n x

ln (20×5) = ln x²

ln(100) = lnx²

100 = x²

x = +/- 10

Since logs of negative numebrs don't exist, we reject -10

3 0
2 years ago
Read 2 more answers
According to the manufacturer, about 26% of sour candy in a package of Sandy's Sours are grape. What is the probability that the
Leto [7]

Answer:


Step-by-step explanation:

We assume that there are 100 sour candies, Thus-  

26 % of candy are grape implies that 26% of 100 candies are grape that is equal to 26

Now remaing candies that are not grape are 100-26 = 74

Based on the rule of multiplication:

P(A ∩ B) = P(A)/ P(B|A)  

 In the beginning, there are 26grape candies, probability of choosing first grape candy =  26C1 = 26

 After the first selection, we replace the selected grape candy so there are still 100 candies in the bag P(B|A) =  100C3 = 100 x 99 x 98 x 97!/3! X 97!

= 50 x 33 x 98

So probability =1/ 50 x 33 x 98 x 26

= 1/4204200



8 0
2 years ago
Read 2 more answers
Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The
abruzzese [7]

The question is incorrect.

The correct question is:

Three TAs are grading a final exam.

There are a total of 60 exams to grade.

(c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

Answer: 60!/(25!20!15!)

Step-by-step explanation:

The number of ways of arranging n unlike objects in a line is n! that is ‘n factorial’

n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

The number of ways of arranging n objects where p of one type are alike, q of a second type are alike, r of a third type are alike is given as:

n!/p! q! r!

Therefore,

The answer is 60!/25!20!15!

6 0
2 years ago
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