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Talja [164]
3 years ago
8

PLEASE HELP: Simplify (B^9)^3

Mathematics
2 answers:
Misha Larkins [42]3 years ago
8 0

Answer:

b^27, so in other words C.

Step-by-step explanation:

Molodets [167]3 years ago
8 0

Answer:

b^27

Step-by-step explanation:

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Answer:

7-3.182 \frac{13.638}{\sqrt{4}}= -14.698

7+3.182 \frac{13.638}{\sqrt{4}}= 28.698

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150  

Weight after diet 170 130 215 152

We define the random variable D = before-after and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

And the deviation with:

s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}

And after replace we got:

\bar D= 7, s_d = 13.638

And the confidence interval for this case would be given by:

\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1= 4-1=3

For the 95% of confidence the value for the significance is \alpha=0.05 and the critical value would be t_{\alpha/2}= 3.182. And replacing we got:

7-3.182 \frac{13.638}{\sqrt{4}}= -14.698

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Step-by-step explanation:

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3 years ago
The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l
mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: (\frac{n+1}{2}) th\ term

Putting n=21

(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

<u>For First Half:</u>

98, 100, 101, 102, <u>108, 109</u>, 111,118, 129, 132

Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5

<u>For Second Half:</u>

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Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151

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IQR = Q_3-Q_1\\= 151-108.5\\=42.5

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

  • brainly.com/question/10940255
  • brainly.com/question/10941043

#LearnwithBrianly

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Answer:

what grade

Step-by-step explanation:

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slope = 7 passing thru (0,1)

equation: y = 7x+1

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