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3 years ago

Please help! Correct answer only!

Mathematics

1 answer:

elena55 [62]3 years ago

5 0

Answer:

<em>" Expected Payoff " - $ 12.30</em>

Step-by-step explanation:

Consider the steps below;

$Total Number of Tickets - 200 Tickets,\\Tickets Entered - 1 Ticket,\\\\Proportion - 1 / 200\\\\Amount Won ( First ) = 980 Dollars,\\Amount Won ( Second ) = 950 Dollars,\\Amount Won ( Third ) = 270 Dollars,\\Amount Won ( Fourth ) = 260 Dollars\\\\Proportionality ( First ) - 1 / 200 = x / 980,\\Proportionality ( Second ) - 1 / 200 = x / 950,\\Proportionality ( Third ) - 1 / 200 = x / 270,\\Proportionality ( Fourth ) - 1 / 200 = x / 260,\\\\1 / 200 = x / 980,\\200 * x = 980,\\x = 4.90,\\\\$

$1 / 200 = x / 950,\\200 * x = 950,\\x = 4.75,\\\\1 / 200 = x / 270,\\200 * x = 270,\\x = 1.35,\\\\1 / 200 = x / 260,\\200 * x = 260,\\x = 1.3$

$Conclusion ; " Expected Payoff " = 4.90 + 4.75 + 1.35 + 1.3 = 12.30$

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The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

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After dividing and simplifying, we get:

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Now, finding the derivative of T(u), we get:

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After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

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Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

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Learn more about the vector here:

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