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3 years ago

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[infinity] n = 1 3n + 14−n Determine whether the geometric series is convergent or divergent. Justify your answer. Converges; th

e series is a constant multiple of a geometric series. Converges; the limit of the terms, an, is 0 as n goes to infinity. Diverges; the limit of the terms, an, is not 0 as n goes to infinity. Diverges; the series is a constant multiple of the harmonic series. If it is convergent, find the sum. (If the quantity diverges, enter DIVERGES.)

2 answers:

siniylev [52]3 years ago

8 0

jwbgfkrshfukhuekhfkuhkufz

Damm [24]3 years ago

3 0

Answer:

n = 14/11

Step-by-step explanation:

n =13n +14 - n

=12n + 14

n - 12n=14

-11n = 14

divide both sides by - 11

i think

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Answer to three friends each create 4 bags of starter bread dough. After ten days, each of those four bags is then divided into

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<span>To begin, each of the friends have made 4 bags of dough. This gives 12 bags at the outset. After 10 days, all 12 bags are made into (4*12) bags, or 48 bags. After 20 days, the 48 bags are made into (48*4), or the required 192 bags. After 20 days, then, the friends have the needed 192 bags.</span>

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3 years ago

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If<br> f(x) = 13x + 1, then<br> f-1(x) =

amid [387]

Answer:

(x-1)/13

Step-by-step explanation:

y = 13x+1

To find the inverse, exchange x and y

x = 13y+1

Solve for y

Subtract 1 from each side

x-1 =13y+1-1

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Divide each side by 13

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3 years ago

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The file "berkeley.dat" contains average yearly temperatures for the cities of Berkeleyand Santa Barbara. Import the data into R

WINSTONCH [101]

Answer:

a. A plot of berkeley vs time and berkeley vs stbarb can be found in the 1st and 2nd image attachments

b. A plot of the residual diagnostics regression of berkeley on time can be found in the 3rd image attachment

c. The residuals obtained after a regression of berkeley and stbarb seem like a reasonable fit as they do not strongly violate the regression assumptions. The ACF plot for the residuals show a correlation at lag 11 that is larger than what is expected if the residuals were independent. However, it is not extremely large and is most likely due to regular variation. A plot of this regression can be found in the fourth image attachment.

d. A plot of the variable berkeley and an ACF plot of the data can be found as the first diagram in the fifth image attachment. The time series here has an increasing trend which means that it could not possibly be stationary. The ACF plot is difficult to interpret since the data is not stationary. It cannot be interpreted as an approximation to the auto-correlation function.

e. A plot of the differenced data and its corresponding ACF plot can be found as the second diagram in the fifth image attachment.

The data here seems to be fairly stationary after differencing with a fairly constant variance and no discernible trend.

f. The formula for the differenced time series would be

$x_1 - x_{t-1} = \beta _1 + w_t - w_{t-1}$

The differenced data has a constant mean. The ACF at lag one is negative and significantly outside the confidence intervals. The other lags show no or weak dependency

Step-by-step explanation:

I have attached images of this solution to represent the various constraints plotted against each other.

a. The plots of berkeley vs time and berkeley vs stbarb are contained in the first and second image attachments.

b. The residual diagnostics plots of a regression of berkeley on time(including the ACF) are contained in the third image attachment.

Inferences can be made thus:

Residual standard error: 0.4539 on 102 degrees of freedom

Multiple R-Squared: 0.4015

Adjusted R-Squared: 0.3956

F-Statistic: 68.43 on 1 and 102 degrees of freedom

The F test therefore indicates a strong relationship

c. The residual diagnostics plots of a regression of berkeley on stbarb(including the ACF) are contained in the fourth image attachment.

Inferences can be made thus:

Residual standard error: 0.5221 on 102 degrees of freedom

Multiple R-Squared: 0.2079

Adjusted R-Squared: 0.2001

F-Statistic: 26.77 on 1 and 102 degrees of freedom

The residuals here do not strongly violate the regression assumptions. The ACF plot for the residuals show a correlation at lag 11 that is larger than what is expected if the residuals were independent. However, it is not extremely large and is most likely due to regular variation.

d. A plot of the variable berkeley and an ACF plot of the data is contained as the first diagram in the fifth image attachment.

The time series here has an increasing trend which means that it could not possibly be stationary. The ACF plot is difficult to interpret since the data is not stationary. It cannot be interpreted as an approximation to the auto-correlation function.

e. A plot of the differenced data and its corresponding ACF plot is find as the second diagram in the fifth image attachment.

The data here seems to be fairly stationary after differencing with a fairly constant variance and no discernible trend.

f. The model after the differencing would be

$x_1 - x_{t-1} = \beta _1 + w_t - w_{t-1}$

The differenced data has a constant mean. This corresponds to the ACF plot in the previous question. The ACF at lag one is negative and significantly outside the confidence intervals. The other lags show no or weak dependency

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4 years ago

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lapo4ka [179]

Answer:

There are no common factor between the two or its just 1.

Explanation

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2 years ago

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Mice21 [21]

Answer:

Common difference(d) $a_{52}$

(21) -10 -548

(22) -7 -323

(23) 10 547

(24) -100 -5118

Step-by-step explanation:

Let the common difference be denoted by 'd'.

Also the nth difference of an arithmetic sequence is given by: $a_{n}=a_{1}+(n-1)\times d$

(21)

We are given a recursive formula as:

$a_{n}=a_{n-1}-10$

The first term is given by:

$a_{1}=-38$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-10$

The nth term of an arithmetic sequence is given by:

$a_{n}=a_{1}+(n-1)\times d$

Here $a_{1}=-38$ and $d=-10$ .

Hence, $a_{52}=-38+(52-1)\times (-10)$

Hence, $a_{52}=-548$

(22)

$a_{n}=a_{n-1}-7$

$a_{1}=34$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-7$

Here $a_{1}=34$ and $d=-7$ .

Hence, $a_{52}=34+(52-1)\times (-7)$

Hence, $a_{52}=-323$

(23)

$a_{n}=a_{n-1}+10$

$a_{1}=37$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=10$

Here $a_{1}=37$ and $d=10$ .

Hence, $a_{52}=37+(52-1)\times (10)$

Hence, $a_{52}=547$

(24)

$a_{n}=a_{n-1}-100$

$a_{1}=-18$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-100$

Here $a_{1}=-18$ and $d=-100$ .

Hence, $a_{52}=-18+(52-1)\times (-100)$

Hence, $a_{52}=-5118$

5 0

4 years ago