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ella [17]
2 years ago
12

If 10 cheese slices cost $2.48 and if 12 cheese slices cost 3.18 which one would be the best to get

Mathematics
2 answers:
vlabodo [156]2 years ago
7 0
Divide 2.48 by 10 to find the price of 1 slice. = 0.248
Do the same with the other price.
3.18 / 12 = 0.265

The best one to get would be the 10 cheese slices because a single slice is cheaper than the single slices for the 12 slices.


hope this helped! (:
Elenna [48]2 years ago
6 0
You would rather want 10 cheese slices, because the unit rate for the 10 slices is 0.248. The unit rate for 12 slices is 0.265.
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(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

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According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

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Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

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The null hypothesis is:

<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

For 5% level of significance the two tailed critical value of <em>z</em> is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct the 95% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

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