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3 years ago

CAN SOMEONE HELP ME PLEASE

Mathematics

1 answer:

klasskru [66]3 years ago

6 0

The correct answer is A.

We can tell this because when you write two triangles as congruent, the parts must match up in order. Since AC are the first and last terms in the triangle, we must have the first and last from the second. D and F are exactly that, so DF is congruent to AC.

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Is it a function????

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3 years ago

Help please! Greatly appreciated if you do!

Kisachek [45]

Answer:

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2 years ago

Read 2 more answers

In a one-month period, a convenience store

Vera_Pavlovna [14]

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2 years ago

12) Which polynomial function has a leading coefficient of 2, root-5 with multiplicity 3, and root 12 with multiplicity 1?

gogolik [260]

Answer:

Step-by-step explanation:

5 0

2 years ago

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

2 years ago