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motikmotik
3 years ago
9

What do you know about porter 1980 and purchase management model, explain?

Mathematics
1 answer:
pogonyaev3 years ago
3 0


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The Academy of Management Journal Vol. 27, No. 3, Sep., 1984 Porter's (1980) Gene...



Porter's (1980) Generic Strategies as Determinants of Strategic Group Membership and Organizational Performance

Gregory G. Dess and Peter S. Davis
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Tessssttttt!!!! ASAP Anwser. 50 COoinnss PLZ Help Can some plz help me with this! I have two attempts on this test. I used one a
skad [1K]

Answer:

i believe its c but im not sure

Step-by-step explanation:

3 0
3 years ago
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Monthly water bills for a city have a mean of $108.43 and a standard deviation of $36.98. Find the probability that a randomly s
GaryK [48]

Answer:

The bill size can be considered usual.

Step-by-step explanation:

Mean (μ) = $108.43

Standard deviation (σ) = $36.98

Now, consider the distribution to be normal distribution :

P(Z=\frac{X-\mu}{\sigma}>\frac{a-\mu}{\sigma})=P(Z>\frac{173-108.43}{36.98})\\\\\implies P(Z>1.75)

Now, finding values of z-score from the table. We get,

P(Z > 1.75) = 0.9599

⇒ 95.99%

So, only 4.01 % of the people in the city wastes water.

Hence, the bill size can be considered usual.

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3 years ago
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What is 18% of 50?<br> _90<br> _50<br> _36<br> _9<br> _18
FrozenT [24]

Answer:

9

Step-by-step explanation:

percentages are reversable. like instead of 18% of 50 you could reverse it and say what is 50% of 18. which is 9.

5 0
3 years ago
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Hilda draws a card from a deck of cards and it is a seven. The probability that it is the 7 of spades is Select one: O a. 1/3 O
Schach [20]

Answer:b

Step-by-step explanation:

Given

Hilda draws a card from a deck of cards and it is seven.

It is certain that card is 7 so there a 4 cards which can be 7

i.e. 7 of hearts,7 of spades,7 of club,7 of diamond

So probability that it is the 7 of spades is

\frac{1}{4}=0.25

8 0
3 years ago
How do I solve question 6 through 8?<br> Solve for me
rewona [7]

The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

<h3>How to determine the functions?</h3>

A quadratic function is represented as:

y = a(x - h)^2 + k

<u>Question #6</u>

The vertex of the graph is

(h, k) = (-1, 2)

So, we have:

y = a(x + 1)^2 + 2

The graph pass through the f(0) = -2

So, we have:

-2 = a(0 + 1)^2 + 2

Evaluate the like terms

a = -4

Substitute a = -4 in y = a(x + 1)^2 + 2

y = -4(x + 1)^2 + 2

<u>Question #7</u>

The vertex of the graph is

(h, k) = (2, 1)

So, we have:

y = a(x - 2)^2 + 1

The graph pass through (1, 3)

So, we have:

3 = a(1 - 2)^2 + 1

Evaluate the like terms

a = 2

Substitute a = 2 in y = a(x - 2)^2 + 1

y = 2(x - 2)^2 + 1

<u>Question #8</u>

The vertex of the graph is

(h, k) = (1, -2)

So, we have:

y = a(x - 1)^2 - 2

The graph pass through (0, -3)

So, we have:

-3 = a(0 - 1)^2 - 2

Evaluate the like terms

a = -1

Substitute a = -1 in y = a(x - 1)^2 - 2

y = -(x - 1)^2 - 2

Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

Read more about parabola at:

brainly.com/question/1480401

#SPJ1

5 0
1 year ago
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