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3 years ago

87g of sugar is needed to make 10 cakes. How much sugar is needed for 7 cakes?

2 answers:

nataly862011 [7]3 years ago

7 0

87g divided by 10 = 8.7
8.7 times 7 = 60.9
60.9g of sugar is needed to make 7 cakes
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ddd [48]3 years ago

6 0

Answer:

7 cake needs sugar = 60 .9 grams.

Step-by-step explanation:

Given : 87g of sugar is needed to make 10 cakes.

To find : How much sugar is needed for 7 cakes.

Solution : We have given

10 cakes needs sugar = 87 g .

1 cake needs sugar = $\frac{87}{10}$ = 8. 7

7 cake needs sugar = 8 .7 * 7 = 60 .9 g

7 cake needs sugar = 60 .9 grams.

Therefore, 7 cake needs sugar = 60 .9 grams.

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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".

$\bar X$ represent the sample mean

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$\sigma$ represent the population standard deviation

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Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

For this case we can calculate the mean like this:

$\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

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$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

Part b

For this case is the sample size is doubled the margin of error would be:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

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