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Law Incorporation [45]
2 years ago
11

the catalogue price of an article is rs.50000 and a manufacturer sells it to the distributor at a discount of 20% of the catalog

ue price.The distributor sells it to the retailer at a discount of 10%of catalogue price. A. what is the profit made by the retailer if he sells the article to the customer at the catalogue price? B.What profit is made by the manufacturer if the catalogue price is 50% above the manufacturing price?​
Mathematics
1 answer:
timofeeve [1]2 years ago
7 0

Answer:

A. what is the profit made by the retailer if he sells the article to the customer at the catalogue price?

10000

B.What profit is made by the manufacturer if the catalogue price is 50% above the manufacturing price?

25000

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Answer:

slope is zero

Step-by-step explanation:

it is a horizontal line which means the slope all points that lie on the line have a -coordinate

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3 years ago
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Use the distributive property to solve the equation 28-(3x+4)=2(x+6)+x. Show your work.
Gre4nikov [31]

Answer:

x=2

Step-by-step explanation:

28−(3x+4)=2(x+6)+x

We need to use the distributive property for the right side.

28−3x−4=(2)(x)+(2)(6)+x)

28−3x−4=2x+2+x

−3x+24=3x+12

From here we need to subtract 3x from both side.

−3x+24−3x=3x+12−3x

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Transfer +24 on the right side.

6x=12−24

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Finally, divide both sides by −6

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2 years ago
A sweater was on sale at 40% off the regular price. Ella saved 20$ by buying the sweater on sale. What was the regular price of
barxatty [35]

Answer:

the answer is $50 for the full price of the sweater

Step-by-step explanation:

if you know $20 is 40% what is 20% it is $10 and then multiple it by 5 because  20 times 5 is 100 and you get 50

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3 years ago
Which expresion does NOT equal the length of one of these pieces in feet. Plz help me i really need it.
Troyanec [42]

Answer: J

Step-by-step explanation:

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The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho
vitfil [10]

Answer:

Therefore, the sampling distribution of \bar{x} is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean \mu is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is  

\sigma = 12.6 \: hours

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

n = 250

The sample mean is  

\bar{x} = 9 \: hours  

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

 \mu = \bar{x} = 9 \: hours

The sample standard deviation would be  

$ s = {\frac{\sigma}{\sqrt{n} }  $

Where   is the population standard deviation and n is the sample size.

$ s = {\frac{12.6}{\sqrt{250} }  $

s = 0.7969 \: hours

Therefore, the sampling distribution of \bar{x} is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

\text {confidence interval} = \mu \pm MoE\\\\

Where the margin of error is given by

$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sampling size, s is the sample standard deviation and  is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\

So the required 95% confidence interval is

\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569

The 95% interval estimate of the population mean \mu is

LCL = 7.431 hours to UCL = 10.569 hours

8 0
3 years ago
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