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Lera25 [3.4K]
3 years ago
15

A particular fruit's weights are normally distributed, with a mean of 405 grams and a standard deviation of 19 grams. If a fruit

is picked at random then 17% of the time, its weight will be greater than how many grams?
Mathematics
1 answer:
cluponka [151]3 years ago
8 0

Answer:

413.126

Step-by-step explanation:

Given a normal distribution :

Mean (m) = 405

Standard deviation (s) = 19

Weight of fruit will be greater than how many grams (x) if picked 17% of the time:

P(Z > x) = 0.17 ;

The Zscore for P(Z > x) = 0.17 equals 0.954 (Z probability calculator)

Zscore = (x - m) / s

0.954 = (x - 405) / 19

0.954 * 19 = x - 405

18.126 = x - 405

x = 18.126 + 405

x = 423.126

Weight will be greater than 423.126

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Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

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Following the pattern, we can see that for f^{n}(x),

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f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

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