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sergey [27]
4 years ago
13

a school district requires all graduating seniors to take a mathematics test. This year, the rest scores were approximately norm

ally distributed for the 1208 graduating seniors. This mean score on the test was a 74 and the standard deviation was 11. what percent of the graduating seniors had a test above 85? Express your answers to the nearest percent.
Mathematics
1 answer:
Klio2033 [76]4 years ago
3 0
A score of 85 would be 1 standard deviation from the mean, 74.  Using the 68-95-99.7 rule, we know that 68% of normally distributed data falls within 1 standard deviation of the mean.  This means that 100%-68% = 32% of the data is either higher or lower.  32/2 = 16% of the data will be higher than 1 standard deviation from the mean and 16% of the data will be lower than 1 standard deviation from the mean.  This means that 16% of the graduating seniors should have a score above 85%.
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Debora [2.8K]

Answer:

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Step-by-step explanation:

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2 years ago
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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

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Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
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Natasha_Volkova [10]

Answer:3x???

Step-by-step explanation:

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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Harrizon [31]

Answer:

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