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kow [346]
3 years ago
6

PLEASE HELP

Mathematics
1 answer:
bekas [8.4K]3 years ago
4 0
The answer should be 50x+10=240
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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

4 0
3 years ago
Helppppp please nowwww helpppp
Fantom [35]

Answer:

B

Step-by-step explanation:

m\angle ABZ+m\angle ZBC=m\angle ABC \\ \\ 12x+55=16x+15 \\ \\ 4x=40 \\ \\ x=10 \\ \\ \therefore m\angle ABZ=120^{\circ}

8 0
2 years ago
The time it takes for a statistics professor to mark a single midterm test is normally distributed with a mean of 4.8 minutes an
Bess [88]

Answer:

Step-by-step explanation:

SD of mean = 2.3/\sqrt{55} = 0.418

5 hours = 300 minutes

300 minutes/55 students = 5.45 min/student

The problem asks what is the probability he needs more than an average of 5 min/student

P(XBAR < 5.0) = P((XBAR - 4.8) / 0.418) < (5.0 - 4.8) /0.418) = P(Z < 0.478) = 0.8893        

P(XBAR > 5.0) = 1 - 0.8893 = 0.1107

7 0
3 years ago
How can you determine if information on website is relevant to your topic
GuDViN [60]
You need to look at website see what your witting about a good way to if the website has a phone number call and find on on your topic and ask questions about it
8 0
3 years ago
What would this problem be factored to ? a^2-b^2
frozen [14]
The answer is
(a-b)(a+b)
7 0
4 years ago
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