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love history [14]
4 years ago
6

A cell phone company says that a fully charged battery lasts 14 hours. In reality, each battery behaves a little differently. Fo

r a large quantity of fully charged batteries, the mean length of time that batteries last is 14 hours, with a standard deviation of 1 hour. Assume that the lengths of time that battery charges last can be modeled by a normal distribution.
What percent of fully charged cell phone batteries would be expected to last longer than 14 hours?
Mathematics
1 answer:
gregori [183]4 years ago
8 0

Answer:

The percentage is 50%

Step-by-step explanation:

From the question we are told that  

     The  mean is  \mu  =  14 \  hours

     The  standard deviation is \sigma =  1 \  hour

Generally on the normal distribution curve the mean divides the curve into  two. with the half to the left of the mean (50% to the left )representing the percentage of the batteries that will last less than 14 hours and  the area under the curve to the right (50% to the right)representing the percentage of batteries that will last longer than mean (14 hours )

Hence the percentage of the cell phone batteries that will be expected to last longer than 14 hours is 50%

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John looked at his bank account and noticed that he had $2,029.90 in his account. All that he could remember is that he created
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We will have that for the previous year, he had in the account:

m_{y6}=2029.90-\frac{2029.90\cdot6}{100}\Rightarrow m_{y6}=1908.106

For the year previous to that one, he had:

m_{y5}=1908.106-\frac{1908.106\cdot6}{100}\Rightarrow m_{y5}=1793.61964

For the previous year to that, he had:

m_{y4}=1793.61964-\frac{1793.61964\cdot6}{100}\Rightarrow m_{y4}=1686.002462

For the year previous to that one, he had:

m_{y3}=1686.002462-\frac{1686.002462\cdot6}{100}\Rightarrow m_{y3}=1584.842314

For the second year that the money was in the account, there was:

m_{y2}=1584.842314-\frac{1584.842314\cdot6}{100}\Rightarrow m_{y2}=1489.751775

And the orinial ammount of money that was in the account was:

m_{y1}=1489.751775-\frac{1489.751775\cdot6}{100}\Rightarrow m_{y1}=1400.366669

From this, we know that Jhon had originally $1400.37 in the bank account.

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