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MAVERICK
2 months ago
7

Identify the horizontal asymptote of each graph. t(x) = 67 y=0 y=1 y=6

Mathematics
2 answers:
PIT_PIT [12.4K]2 months ago
6 0

Respuesta:

Primer gráfico (mostrado arriba): A) y = 0

Segundo gráfico: y = -3

tester [12.3K]2 months ago
3 0

Respuesta:

t(x) = 6^x.   El primer gráfico es y=0

t(x) = 5^x -3                  El segundo gráfico es y= -3

Explicación paso a paso:

Correcto en el límite.

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You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $1; if
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Step-by-step explanation:

The scenario is as follows: You and a friend participate in a game involving tossing a fair coin.

The sample space for tossing two coins is {TT, HT, TH, HH}

Let Y represent the earnings from one round of the game.

If both faces are heads, you win $1; therefore, P(Y=1)=P(TT)=\dfrac{1}{4}=0.25

You win $6 if both faces are heads, so P(Y=6)=P(HH)=\dfrac{1}{4}=0.25

If the faces do not match, you lose $3 which means P(Y=1)=P(TH, HT)=\dfrac{2}{4}=0.50

To find the expected value to win: E(Y)=\sum_{i=1}^{i=3} y_ip(y_1)

=1\times0.25+6\times0.25+(-3)\times0.50=0.25

Thus, the mean of Y: E(Y)= $0.25

E(Y^2)=\sum_{i=1}^{i=3} y_i^2p(y_i)\\\\=1^2\times0.25+6^1\times0.25+(-3)^2\times0.5\\\\=0.25+1.5+4.5=6.25

Variance = E[Y^2]-E(Y)^2

=6.25-(0.25)^2=6.25-0.0625=6.1875\approx6.19

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2 months ago
Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t
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Answer:   a) \bold{\dfrac{3}{16}}     b) \bold{\dfrac{1}{36}}

Step-by-step explanation:

a) To achieve an even total, there are 3 possible combinations:

1) Even, Even, Even, Even     \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

2) Even, Even, Odd, Odd   \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

3) Odd, Odd, Odd, Odd   \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

Order is irrelevant

Summing these yields your final result: \dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}

b) If one die shows a 2 and another a 3, while the remaining two can show any digits, there’s only one way to get a 2, one way for a 3, and six potential numbers for each of the other two dice.

\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}

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