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Taya2010 [7]
3 years ago
15

A piece of fabric costs $5.29 a yard which is estimated cost of 16.3 yards of fabric

Mathematics
2 answers:
Cloud [144]3 years ago
5 0

Answer:

your estimate would be $86.22

Step-by-step explanation:

bekas [8.4K]3 years ago
5 0

= $5.29 per yard × 16.3 yards= $86.227

ANSWER: The exact answer is $86.227 for 16.3 yards.  Not knowing the answer choices, you could estimate to $86.00, you could estimate to $86.23, you could even estimate to $90.00.

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Please help I don’t think I remember this
nikdorinn [45]

Answer:

im not sure if this is right

Step-by-step explanation:

demension 1: 9 X 10      demension 2: 10 x 4

4 0
3 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

4 0
3 years ago
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saveliy_v [14]

Answer:

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3 years ago
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yuradex [85]

Answer:

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Step-by-step explanation:

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The numbers of students in the four
Ludmilka [50]

The total number of students in the four classes is 100 students.

<h3>How to compute the value?</h3>

From the information, the numbers of students in the four sixth-grade classes at Northside School are 26, 19, 34, and 21.

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Learn more about numbers on:

brainly.com/question/24644930

#SPJ1

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