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

1 answer:

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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Someone help me please:)

IRISSAK [1]

Im not that good with these types of problems but i think it is b if my math is correct.

6 0

3 years ago

Can someone pls help me 2-7 thank u :)

kolbaska11 [484]

2. 27
3. 13
4. 35
6. 22
7. 60

I was unable to help with 5 because of the plot box!! Hope this helps

8 0

3 years ago

Read 2 more answers

Plz help! read the instructions and write a proper answer and if you don't i will report you! And the first one to answer right

ahrayia [7]

Answer:

1 = 0.61, 2 = 1.13

Step-by-step explanation:

The first is 25 + 25 + 10 + 1. This is 61 cents which is 0.61 of 1 dollar. The second is 100 + 10 + 1 + 1 + 1. This is equal to 113, 0r 1.13

BRAINLIEST PLEASE

3 0

2 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$