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2 years ago

Which sequence can be defined by the recursive formula f (1) = 4, f (n 1) = f (n) – 1.25 for n ≥ 1?

2 answers:

7 0

F(1) = 4; f(n+1) = f(n) - 1.25

f(1) = 4

f(2) = f(1) - 1.25 = 4 - 1.25 = 2.75

f(3) = f(2) - 1.25 = 2.75 - 1.25 = 1.50

f(4) = 1.50 - 1.25 = 0.25

f(5) = 0.25 - 1.25 = -1

f(6) = -1 -1.25 = -2.25

Sequence: 4, 2.75, 1.50, 0.25, -1, -2.25, ...

Leona [35]2 years ago

7 0

Answer:

C).4, 2.75, 1.5, 0.25, –1, . . .

Step-by-step explanation:

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2 years ago

You are giving the following amounts: $190,258.50; $152,698.00; $122,753.00; $220,523.00; $231,951.00. What is the average of th

blondinia [14]

Answer:

Hey!

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Step-by-step explanation:

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4 0

2 years ago

The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in

Marina86 [1]

Answer:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

$\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3$

We can express this formula like this:

$\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

$\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

If we operate and we take out the 1/4 as a factor we got this:

$\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}$

We can cancel $n^2$ and we got

$\lim_{n\to\infty} \frac{(n+1)^2}{n^2}$

We can reorder the terms like this:

$\lim_{n\to\infty} (\frac{n+1}{n})^2$

We can do some algebra and we got:

$\lim_{n\to\infty} (1+\frac{1}{n})^2$

We can solve the square and we got:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

3 0

2 years ago

Select all the properties that are used below to solve 7y − 15 = −29. 7y – 15 + 15 = –29 + 15 7y = –14 7y7 = –147 y = –2 A. Iden

lukranit [14]

B.addition property of multiplication

D.inverse property of multiplication

E.commutative property of addition

3 0

2 years ago

According to a human modeling project, the distribution of foot lengths of women is approximately Normal with a mean of 23.3 ce

Yakvenalex [24]

Answer:

26.11% of women in the United States will wear a size 6 or smaller

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 23.3, \sigma = 1.4$

In the United States, a woman's shoe size of 6 fits feet that are 22.4 centimeters long. What percentage of women in the United States will wear a size 6 or smaller?

This is the pvalue of Z when X = 22.4. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{22.4 - 23.3}{1.4}$

$Z = -0.64$

$Z = -0.64$ has a pvalue of 0.2611

26.11% of women in the United States will wear a size 6 or smaller

8 0

2 years ago