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solong [7]
3 years ago
7

Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 skat

eboards and a standard deviation of 55 skateboards.
(a) On what percent of the days last year did the factory produce 20,555 skateboards or fewer?
(b) On what percent of the days last year did the factory produce 20,610 skateboards or more?
(c) On what percent of the days last year did the factory produce 20,445 skateboards or fewer?
Mathematics
1 answer:
alisha [4.7K]3 years ago
8 0

Answer:

This can be solved by using the empirical rule for a normal distribution.

Step-by-step explanation:

A. The number of skateboards given is one standard deviation above the mean. Approximately 68% of the data points lie within the range plus and minus one standard deviation of the mean. Therefore the required percentage is:

68 + 16 = 84%.

B. The given number of skateboards is two standard deviations above the mean. Approximately 95% of the data points lie within the range plus and minus two standard deviations of the mean. Therefore the required percentage is:

5/2 = 2.5%

C.The given number of skateboards is one standard deviations below the mean. Therefore the required percentage is:

16%.

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a\times (-a)\times 13\times a\times (-a)\times 13 can be written in power notation as a^{4}\times 13^{2}

Step-by-step explanation:

The given expression

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Writing a\times (-a)\times 13\times a\times (-a)\times 13 in power notation:

Let

a\times (-a)\times 13\times a\times (-a)\times 13

= [13\times13][(a\times (-a)\times a\times (-a)]

As

13\times13 = 13^{2} , a\times a = a^{2} , (-a)\times (-a) = (-a)^{2}

So,

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So,

=[13^{2}][a^2\times a^2]

As ∵a^{m} \times a^{n}=a^{m+n}

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As ∵a^{m} \times a^{n}=a^{m+n}

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Therefore, a\times (-a)\times 13\times a\times (-a)\times 13 can be written in power notation as a^{4}\times 13^{2}

<em>Keywords: power notation</em>

<em>Learn more about power notation from brainly.com/question/2147364</em>

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