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3 years ago

A certain type of chocolate bar is sold as follows.

Mathematics

1 answer:

Scrat [10]3 years ago

5 0

Answer:

Step-by-step explanation:

You would never buy a 10-pack, because two 5-packs cost less.

To buy 3 bars you need a single bar and a pair of single bars for £0.82 + £0.82+ £0.41 = £2.05

To buy 30 bars you need 3 pairs of 5-bar packs = 3×£5.75 = £17.25

To buy the remaining 5 bars you need one 5-bar pack = £3

cost = £2.05 + £17.25 + £3 = £22.30

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15 3/5% in fractions

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3 years ago

Verify the following identity: (csc theta -cot theta )/(sec theta -1) A. sec theta B. cos theta C. cot theta D. sin theta

alexdok [17]

Answer:

C. cot theta

Step-by-step explanation:

(csc theta -cot theta )/(sec theta -1)

csc = 1/ sin

cot = cos / sin

sec = 1 / cos

Let x = theta

(1/ sin x -cos x / sin x )/(1/ cos x -1)

Getting a common denominator in the denominator and combining terms

(1- cos x)/ sinx / ( 1 - cos x) / cos x

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Copy dot flip

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3 0

3 years ago

Sample Size for Proportion As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of gol

Dima020 [189]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.025$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume an estimated proportion of $\hat p =0.5$ since we don't have prior info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56$

And rounded up we have that n=2663

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