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cluponka [151]
3 years ago
14

What is an example of dealer incentives ?

Mathematics
1 answer:
Sholpan [36]3 years ago
5 0
A car manufacturer giving a car to a dealer in an reduced amount then the actual selling amount with less profit , through which the car dealer is profited and stick to the business and increases sales from his side by which also the manufacturer is also benefited by the dealers action as he get increased sales margin.Typically dealer incentives increase marketing to slow-moving stock after initial sales objectives have been met.
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G = (5,-10),(25,-50),(50,100)
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This is the answer not sure if you wanted it graph or not

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benjamin finds 31 pine cones at the park. Together Jenna and Ellen find the same number of pinecones as Benjamin. How many pinec
elixir [45]
Solve Step by Step: 

1) Write the Equation: 31/2 = how many pine cones Jenna and Ellen found.
2) Solve 31/2= 15.5

So each girl found 15 and a half pine cones. Or one girl found 15 and the other found 16. If you want to have each girl find an even amount.
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10/m+n<br> m=-2<br> n=-1<br> what is the answer
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negative 3 and one third

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3 years ago
55 + 54 + x + 74 = 180<br> what would x be?
Allisa [31]
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6 0
3 years ago
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In order to justify this claim for the Ferengi a sample of 16 coils is taken. For this sample set, the mean warp phase flux is f
Kay [80]

Answer:

The responses to these question can be defined as follows:

Step-by-step explanation:

n = 16\\\\  \bar{x}= 73.7\\\\\sigma = 12\\\\a = 0.05\ or\  a = 0.10\\\\H_{o} \ : \mu = 68\\\\H_{a} \ : \mu \neq 68\\\\a = 0.05\\\\

critical values =\pm t0.025,15 = \pm 2.131\\\\

(n-1) = 15^{\circ}\\\\a = 0.10\\\\

critical values= \pm t0.05,15 = \pm 1.753\\\\

(n-1) = 15^{\circ}\\\\

Testing the statistic values:

t = \frac{x-\mu_{0}}{ \frac{s}{\sqrt{n}}}\\\\

  = \frac{73.7-68}{(\frac{12}{\sqrt{16}})}\\\\\  = \frac{5.7}{(\frac{12}{4})}\\\\  = \frac{5.7}{(3)}\\\\  = 1.9\\

Test statistic ta = -1.90\ lies

The critical values\pm t_{0.05,15} =\pm 1.753

It is in the region of dismissal. We dismiss the 10% significant null hypothesis.

t_a = 1.90 \\\\df = 15\\\\a = 0.05\\\\p-value = 076831\\\\

P - value is greater than the level of significance a= 0.05  

Null hypothesis we don't reject. At a 95% level, the claim is justified.

t_a = 1.90\\\\ df = 15\\\\  a = 0.10\\\\p-value = 076831\\\\

P - value below the meaning level a = 0.10, we reject the hypothesis null. At a level of 90% the claim is not justified.

3 0
3 years ago
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