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noname [10]
3 years ago
12

Which measure of center would you use to compare the symmetric populations represented by the box plots?

Mathematics
2 answers:
trapecia [35]3 years ago
4 0

Answer:

Step-by-step explanation:

1. mean 2. mean absolute deviation

Ilya [14]3 years ago
3 0

Answer:

1. mean 2. mean absolute deviation

Step-by-step explanation:

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The diagram shows a translation of quadrilateral abcd <br> What is the symbolic representation
lilavasa [31]

9514 1404 393

Answer:

  D.  (x, y) ⇒ (x +8, y -2)

Step-by-step explanation:

The image is 8 units right and 2 units down from the preimage. That is, each x-coordinate is increased by 8, and each y-coordinate is decreased by 2. The appropriate choice is ...

  (x, y) ⇒ (x +8, y -2)

5 0
2 years ago
Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
3 years ago
Simplify the expression. 10 (n^2+ n) - 6 (n^2 - 2)
Gnom [1K]
<h3>10(n²+n)-6(n²+2)</h3><h3>10n²+10n-6n²-12</h3><h3>10n²-6n²+10n-12</h3><h3>4n²+10n-12</h3>

please mark this answer as brainlist

3 0
3 years ago
OCCHICIS
patriot [66]

Answer:

Witch side is shaded if one is shaded then that one is the answer

Step-by-step explanation:

That’s it really if you need help email me

7 0
3 years ago
What is the balanced equation for the reaction (NH4)2SO4+SrCl2--&gt;
steposvetlana [31]
(NH4)2SO4+SrCl2--><span>(NH4)2Cl2 + SrSO4

The reaction side are both aquas and the product is aquas and a solid precipitate. </span>
7 0
3 years ago
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