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ch4aika [34]
2 years ago
7

A cereal company fills boxes with 16 ounces of cereal. The acceptable percent error in filling a box is 2.5%. Find the least and

the greatest acceptable weights. Show your work please
Mathematics
1 answer:
Ket [755]2 years ago
8 0
If the acceptable percent error is 2.5%, then the amount it can be over or under 16 oz is 0.4oz.
16 + 0.4
16 - 0.4
16.4 is the greatest, 15.6 is the least
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There are 15 people in a party, including Hannah and Sarah. We divide the 15 people into 3 groups, where each group has 5 people
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Answer:15/3

Step-by-step explanation:

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A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w
sveta [45]

Answer:

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

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".  

p_A represent the real population proportion for business  

\hat p_A =\frac{75}{400}=0.1875 represent the estimated proportion for Business

n_A=400 is the sample size required for Business

p_B represent the real population proportion for non Business

\hat p_B =\frac{137}{500}=0.274 represent the estimated proportion for non Business

n_B=500 is the sample size required for non Business

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

7 0
1 year ago
In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going trough point M intersects diagonal AC a
geniusboy [140]

Answer:

BC:BN=8:3

Step-by-step explanation:

ABCD is a trapezoid and there is a point m which belongs to AD such that AM:MD=3:5.Line "l" parallel to AB intersects the diagonal AC at p and BD at N.

Now, we know that the parallel lines divide the transversal into the segments with equal ratio, therefore, BN:NC=AM:MD

But, BC= BN+NC

Therefore, BC:BN=(BN+NC):BN

⇒BC:BN=(3+5):3

⇒BC:BN=8:3


4 0
1 year ago
The diagram shows several planes, lines, and points. Vertical plane P intersects horizontal plane R at line f. Vertical plane T
Rudiy27

Please mark brainliest

Answer:

Line H has points on planes R, T, P

Step-by-step explanation:

I just took the test on Edgenuity 2020. Good luck

5 0
1 year ago
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