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kykrilka [37]
3 years ago
6

Daniel bought a van for Rs258 750.This price included a VAT of 15%.Calculate the price of the van without VAT

Mathematics
2 answers:
ch4aika [34]3 years ago
8 0

Answer:

Step-by-step explanation:

price with vat=258750

VAT %=15%

let original price be x

CP+VAT%*CP= Rs258750

x+15/100*x= Rs258750

100x+15x= Rs258750*100

115x= Rs25875000

x= Rs25875000/115

x= Rs225000

iVinArrow [24]3 years ago
3 0

Step-by-step explanation:

is answer

Daniel bought a van for Rs258 750.This price included a VAT of 15%.Calculate the price of the van without VAT

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8 =g

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3 years ago
Look at this graph.
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Step-by-step explanation:

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3 years ago
Josslyn has nickels and dimes in her pocket. The number of nickels is three more than seven times the number of dimes let d repr
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7d+3Explanation

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Step 1

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3 0
1 year ago
A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

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

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

3 0
3 years ago
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