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bearhunter [10]
2 years ago
13

A. 140° B. 300° C. 220° D. 180° E. 320°

Mathematics
1 answer:
kvasek [131]2 years ago
3 0

Answer:

140

Step-by-step explanation:

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What is the least possible degree of a polynomial that has roots -5,1 + 4i, and -4i?
Oksanka [162]

Without any extra conditions, the answer could be 3, and the simplest polynomial with the given roots would be

(<em>x</em> + 5) (<em>x</em> - (1 + 4<em>i</em> )) (<em>x</em> + 4<em>i</em> )

= <em>x</em> ³ + 4<em>x</em> ² + (11 - 4<em>i</em> ) <em>x</em> + 80 - 2<em>i</em>

<em />

If the polynomial is supposed to have only <em>real</em> coefficients, then any complex roots must occur along with their complex conjugates:

(<em>x</em> + 5) (<em>x</em> - (1 + 4<em>i</em> )) (<em>x</em> - (1 - 4<em>i</em> )) (<em>x</em> + 4<em>i</em> ) (<em>x</em> - 4<em>i </em>)

= <em>x</em> ⁵ + 3<em>x</em> ⁴ + 23<em>x</em> ³ + 133<em>x</em> ² + 112<em>x</em> + 1360

and then the degree would be 5.

6 0
3 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
3 years ago
Geometry help 10 points
Serhud [2]
The answer to the reflection question would be D.
3 0
3 years ago
Read 2 more answers
A factory received a shipment of 11 hammers, and the vendor who sold the items knows there are 2 hammers in the shipment that ar
zhenek [66]

Question:

If a sample of 2 hammer is selected

(a) find the probability that all in the sample are defective.

(b) find the probability that none in the sample are defective.

Answer:

a Pr = \frac{2}{110}

b Pr = \frac{72}{110}

Step-by-step explanation:

Given

n = 11 --- hammers

r = 2 --- selection

This will be treated as selection without replacement. So, 1 will be subtracted from subsequent probabilities

Solving (a): Probability that both selection are defective.

For two selections, the probability that all are defective is:

Pr = P(D) * P(D)

Pr = \frac{2}{11} * \frac{2-1}{11-1}

Pr = \frac{2}{11} * \frac{1}{10}

Pr = \frac{2}{110}

Solving (b): Probability that none are defective.

The probability that a selection is not defective is:

P(D') = \frac{9}{11}

For two selections, the probability that all are not defective is:

Pr = P(D') * P(D')

Pr = \frac{9}{11} * \frac{9-1}{11-1}

Pr = \frac{9}{11} * \frac{8}{10}

Pr = \frac{72}{110}

8 0
2 years ago
3 consecutive odd integers such that the sum of twice the first and three times the second is 55 more than twice the third
den301095 [7]

Answer:

Consecutive odd integers are 19 , 21 & 23

Step-by-step explanation:

Let the first 3 consecutive odd intergers be x , (x + 2) and (x + 4).

According to the question,

2x + 3(x + 2) = 2(x + 4) + 55

=  > 2x + 3x + 6 = 2x + 8 + 55

Eliminating 2x from both the sides,

=  > 3x + 6 = 63

=  > 3x = 63 - 6 = 57

=  > x =  \frac{57}{3}  = 19

So, the consecutive odd integers are = 19 , 21 & 23.

3 0
2 years ago
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