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e-lub [12.9K]
3 years ago
8

A sequence can be generated by using fn = 2f(n-1)+1, where f1 = 4 and n is a whole number greater than 1. What are the first fou

r terms in the sequence?
Mathematics
1 answer:
hoa [83]3 years ago
4 0

Answer:

<em>{9,19,39,79}</em>

Step-by-step explanation:

<u>Recursive Sequences</u>

The recursive sequence can be identified because each term is given as a function of one or more of the previous terms. Being n an integer greater than 1, then:

f(n) = 2f(n-1)+1

f(1) = 4

To find the first four terms of the sequence, we set n to the values {2,3,4,5}

  • For n=2

f(2) = 2f(1)+1

Since f(1)=4:

f(2) = 2*4+1

f(2) = 9

  • For n=3

f(3) = 2f(2)+1

Since f(2)=9:

f(3) = 2*9+1

f(3) = 19

  • For n=4

f(4) = 2f(3)+1

Since f(3)=19:

f(4) = 2*19+1

f(4) = 39

  • For n=5

f(5) = 2f(4)+1

Since f(4)=39:

f(5) = 2*39+1

f(5) = 79

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Answer: 18x2 + 69x + 65

Step-by-step explanation:

(3x + 5) (~6x + 13)

1. 3x(6x + 13) + 5(6x + 13)

2. 18x2 + 39x + 5(6x + 13)

3. 18x2 + 39x + 30 + 65

4. 18x2 + 69 + 65

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At the 2012 Summer Olympic Games in London, in the Men's Shot Put qualifying round, the distances ranged from 17.58 meters to 21
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Answer:

One standard deviation above the mean means a z-score of 1.0. If we put this into the calculator as normalcdf(1,999) = .159

15.9 percent of the athletes would have a distance greater than 20.56 meters.

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.159 * 40 = 6.346 = about 6 athletes

c. about 6 athletes

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As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a
REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105

Distribution of the difference:

p = p_T - p_A = 0.73 - 0.47 = 0.26

s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019

Confidence interval:

Is given by:

p \pm zs

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Lower bound:

p - 1.96s = 0.26 - 1.96*0.019 = 0.223

Upper bound:

p + 1.96s = 0.26 + 1.96*0.019 = 0.297

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

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2 years ago
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