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Alja [10]
3 years ago
14

Stuck on this one please help

Mathematics
1 answer:
Mashutka [201]3 years ago
7 0

Answer:

3

Step-by-step explanation:

3^10= 3^7(x^3)

59049= 2187(x^3)

59049/2187= x^3

27=x^3

Do the square root of 3 to both sides and you end up with x=3.

Hope this helped :)

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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.
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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.

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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.

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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.

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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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