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Llana [10]
3 years ago
5

Find the mean for the following group of data items. 4.1, 8.9, 3.2, 1.9, 7.3, 6.3, 6.7, 8.6, 3.2, 2.3, 5.9 (Round to 3 decimal p

laces as needed.) The mean is
Mathematics
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

The mean is 5.309.

Step-by-step explanation:

Given group of data,

4.1, 8.9, 3.2, 1.9, 7.3, 6.3, 6.7, 8.6, 3.2, 2.3, 5.9,

Sum = 4.1+ 8.9 + 3.2 + 1.9 + 7.3 + 6.3 + 6.7 + 8.6 + 3.2 + 2.3 + 5.9 = 58.4,

Also, number of observations in the data = 11,

We know that,

Mean=\frac{\text{Sum of all observation}}{\text{Total observations}}

Hence, the mean of given data = \frac{58.4}{11}=5.30909\approx 5.309

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Describe one error that was made in the preparation of the
pashok25 [27]

Answer:

  the axis labels are inconsistent with the graph title

Step-by-step explanation:

The independent variable is described as "% protein digested", and the dependent variable is described as "time." The graphed values suggest that these are reasonable descriptors for the data being plotted.

The title, however, says the data points are "percentage digestion per hour". This is in disagreement with the axis labels, and is inconsistent with the shape of the curve. (If the title is to be believed, the digestion rate is such that more than 100% of <whatever> has been digested after 10 hours.)

__

<em>Additional comment</em>

Another error is the vertical axis is graduated as though it were linear, but it is decidedly non-linear. Equivalent distances on the graph are shown for differences of 10%, 5%, 10% and 25%. That is the scale varies by a factor of 5 from one part of the graph to another.

8 0
3 years ago
Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com
Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.9842

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0
3 years ago
One number is three more than the other. their sum is thirty three. find the numbers
adell [148]
The numbers are 15,18
They add to 33 and a difference of 3
3 0
3 years ago
Is an Obtuse triangle a Acute triangle? Yes or no? And why or why not?
777dan777 [17]

Answer:

No, this is because the word obtuse is in the name obtuse triangle as well as it being a whole different angle than an acute angle. Obtuse angles are are greater than 90 degress but less than 180 while the acute angle is less than 90.

Step-by-step explanation:

5 0
3 years ago
drinking 6 fluid ounces of milk provides 202.5 mg of calcium how many fluid ounces of milk are required to provide 52.5 mg of ca
Ymorist [56]

Answer:

2 fluid ounces.

Step-by-step explanation:

202.5 divided by 6 = 33.75 (1 fluid ounce)


33.75 x 2 = 67.5.  (68)  







5 0
3 years ago
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