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vlabodo [156]
3 years ago
14

Suppose N coins lie heads up on a table. In one turn, you can turn over any (N−1) coins. Is it possible that N coins could lay t

ails up in any number of turns, if a N=15 ?
Mathematics
1 answer:
kirza4 [7]3 years ago
6 0

Answer:

The answer is no

Step-by-step explanation:

no

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Find the measure of 2.<br> 63° 117°<br> 62 = [?]<br> Enter
forsale [732]

Answer:

63°

Step-by-step explanation:

Angle <2 and the angle with measure of 63 are alternate exterior angles and has same measurement.

7 0
3 years ago
Find the missing side . round the the nearest tenth
lys-0071 [83]

S\frac{O}{H} C\frac{A}{H} T\frac{O}{A}

Sin=\frac{Opposite}{Hypotenuse} Cos=\frac{Adjacent}{Hypotenuse} Tan=\frac{Opposite}{Adjacent}

7) You have the opposite. You need the adjacent.

Tan=\frac{Opposite}{Adjacent}

Tan57=\frac{12}{x}

x=\frac{12}{Tan57}

x = 7.79289... =

<h2>7.8</h2><h2 />

8) You have the hypotenuse. You need the Opposite.

Sin=\frac{x}{Hypotenuse}

Sin37=\frac{x}{13}

x = sin37 × 13

x = 7.8235... =

<h2>7.8</h2><h2 />

9) You have the hypotenuse. You need the opposite.

Sin=\frac{x}{Hypotenuse}

Sin59=\frac{x}{11}

x = sin59 × 11

x = 9.4288.... =

<h2>9.4</h2><h2 />

10) You have the hypotenuse. You need the opposite.

Sin=\frac{x}{Hypotenuse}

Sin53=\frac{x}{11}

x = sin53 × 11

x = 8.7459.... =

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7 0
3 years ago
Suppose that two teams play a series of games that ends when one of them has won ???? games. Also suppose that each game played
Musya8 [376]

Answer:

(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0

(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0

Step-by-step explanation:

(a) when i = 2, the expected number of played games will be:

E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] =  4p²-4p+2-6p²+6p = -2p²+2p+2.

If p = 1/2, then:

d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.

(b) when i = 3;

E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]

Simplification and rearrangement lead to:

E(X) = 6p⁴-12p³+3p²+3p+3

if p = 1/2, then:

d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10

Therefore, E(X) is maximized.

6 0
3 years ago
I keep getting this question wrong
padilas [110]
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3 0
2 years ago
An inequality is shown below.
Stells [14]
B.
You are looking for something that does not let 25x go over 80.
8 0
3 years ago
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