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3 years ago

Find the derivative of f(x)= e^(4x) + e^-(4x)

Mathematics

2 answers:

ololo11 [35]3 years ago

4 0

Answer:

$f'(x)=4e^{4x} -4e^{-4x}$

Step-by-step explanation:

To find this derivative, we will need to use the chain rule.

As there is a variable in the exponent we can use this formula:

$f'(x)=u'e^u$

In this case, $u=4x$ and $u=-4x$

This means that $u'=4$ and $u'=-4$ respectively

This gives us $f'(x)=4e^{4x} -4e^{-4x}$

xeze [42]3 years ago

4 0

Answer:

Step-by-step explanation:

note : (e^(u(x))' = (u(x))'e^(u(x)

f(x)= e^(4x) + e^-(4x)

f'(x) =4e^(4x) - 4 e^-(4x)

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3 years ago

4. Find x if PQ = RS, PQ = 9x - 7, and RS = 29.

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4 years ago

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2 years ago

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I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.

masya89 [10]

Answer:

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

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3 years ago