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AVprozaik [17]
2 years ago
8

Listed below are the measured radiation emissions (in W/kg) corresponding to cell phones: A, B, C, D, E, F, G, H, I, J, and K re

spectively. The media often present reports about the dangers of cell phone radiation as a cause of cancer. Cell phone radiation must be 1.6 W/kg or less. Find the a. mean, b. median, c. midrange, d. mode for the data. Also complete part e.
1.47 1.46 1.38 0.26 0.57 0.92 0.44 0.67 0.55 0.36 1.56

a. find the mean

b. Find the median.

c. Find the midrange.

d. Find the mode.

e. If you are planning to purchase a cell phone, are any of the measures of center the most important statistic? Is there another statistic that is most relevant? If so, which one?
Mathematics
1 answer:
Valentin [98]2 years ago
6 0

Answer:

the mean would be 1.47 and 1.56

Step-by-step explanation:

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A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. A
Naddik [55]

Answer:

Initial bacterias = 6006000

Altought I believe is safe to assume that the values were 192,000 and 384,000 instead of 192,192,000 and 384,384,000, in that case the initial bacterias is 6000

Step-by-step explanation:

A exponential growth follows this formula:

Bacterias  = C*rⁿ

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Bacterias (55 hours) = 192,192,000

Bacterias (66 hours) = 384,384,000

Bacterias(55hours)=C*r^{{\frac{55-t}{t}}} \\Bacterias (66hours) = C*r^{\frac{66-t}{t}}}

If you divide both you can get the growth rate:

\frac{Bacterias (66hours)}{Bacterias(55hours)}=\frac{C*r^{\frac{66-t}{t}}}{C*r^{{\frac{55-t}{t}}}} \\\frac{384,384,000}{192,192,000} =r^{\frac{66-t}{t} -\frac{55-t}{t} } \\2 =r^{\frac{11}{t}}

So with that r = 2 and each time interval correspond to 11 years

Then replacing in one you can get the initial amount of C

Bacterias (55hours)=C*2^{\frac{55-11}{11} } 192,192,000 = C*32\\C= 6006000

7 0
3 years ago
Find the rate and unit rate of these please!
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2(4x+7)\:ft

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