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3 years ago

An item is sold for £305. If this includes a mark up of 7%. Find the buying in price

Mathematics

1 answer:

Lady_Fox [76]3 years ago

4 0

Answer:

Let the original price be 'x'

7/100 * x + x = 305

(7x+100x)/100 = 305

107x = 305 * 100

107x= 30500

x= 30500/107

x= 285.04

Therefore, the original buying price is £ 285.04

Hope you understood......

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2 years ago

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3 years ago

What is the answer and how to do this?

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Step-by-step explanation:

Given :

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2 years ago

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo

never [62]

Answer:

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that $\pi = 0.08$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}$

$0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}$

$\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2$

$n = 996.3$

Rounding up:

A sample of 997 is needed.

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3 years ago