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mr_godi [17]
4 years ago
5

Write the standard form and expanded form Five and nine tenths

Mathematics
2 answers:
astraxan [27]4 years ago
8 0
Standard form- 5 9/10
torisob [31]4 years ago
7 0
Standard: 5.9 or 5 9/10
Expanded: 5+0.9
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What is true volume of the solids
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3 years ago
What would be the approximate 95% confidence interval for the mean number of ounces of catchup bottle in the sample
Alex777 [14]

Answer:

The 95% confidence interval for the mean number of ounces of ketchup bottle is (23.8, 24.2).

Step-by-step explanation:

The complete question is:

Suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.8 ounces. If you took a sample of 49 bottles of ketchup, what would be the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample?

Solution:

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

CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}

The information provided is:

\bar x=24\\\sigma=0.8\\n=49\\\text{Confidence Level}=95\%

The critical value of <em>z</em> for 95% confidence level is:

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

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the mean number of ounces of ketchup per bottle as follows:

CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}

     =24\pm1.96\cdot \frac{0.80}{\sqrt{49}}\\\\=24\pm 0.224\\\\=(23.776, 24.224)\\\\\approx (23.8, 24.2)

Thus, the 95% confidence interval for the mean number of ounces of ketchup bottle is (23.8, 24.2).

3 0
3 years ago
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