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Margaret [11]
3 years ago
11

Please help !!! don’t take my points please

Mathematics
1 answer:
Zanzabum3 years ago
3 0
I think the first spinner in the first picture is a 75% chance of landing on yellow
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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

7 0
4 years ago
List possible similarities or differences between photographers today and those in the past to support your argument.
Artyom0805 [142]

Step-by-step explanation:

the past photographers used to setup time while the present photographers and even a selfie's

3 0
3 years ago
Emily has a bag of 20 fruit flavour sweets. 7 of the sweets are strawberry flavour, and 11 are lime flavour.
ycow [4]

Answer:

e

Step-by-step explanation:

4 0
3 years ago
In the diagram below, assume that all points are given in rectangular coordinates. Determine the polar coordinates for each poin
Korolek [52]

Step-by-step explanation:

We have cartisean points. We are trying to find polar points.

We can find r by applying the pythagorean theorem to the x value and y values.

r  {}^{2} =  {x}^{2}  +  {y}^{2}

And to find theta, notice how a right triangle is created if we draw the base(the x value) and the height(y value). We also just found our r( hypotenuse) so ignore that. We know the opposite side and the adjacent side originally. so we can use the tangent function.

\tan(x)  =  \frac{y}{x}

Remeber since we are trying to find the angle measure, use inverse tan function

\tan {}^{ - 1} ( \frac{y}{x} )  =

Answers For 2,5

{2}^{2}  +  {5}^{2}  =  \sqrt{29}  = 5.4

So r=sqr root of 29

\tan {}^{ - 1} ( \frac{5}{2} )  = 68

So the answer is (sqr root of 29,68).

For -3,3

{ -3 }^{2}  +  {3}^{2}  =  \sqrt{18}  = 3 \sqrt{2}

\tan {}^{ - 1} ( \frac{3}{ - 3} )  =  - 45

Use the identity

\tan(x)  =  \tan(x + \pi)

So that means

\tan(x)  = 135

So our points are

(3 times sqr root of 2, 135)

For 5,-3.5

{5}^{2}  +  {3.5}^{2}  =  \sqrt{37.25}

\tan {}^{ - 1} ( \frac{ - 3.5}{ - 5} )  = 35

So our points are (sqr root of 37.25, 35)

For (0,-5.4)

{0}^{2}  +  { - 5.4}^{2}  = \sqrt{}  29.16 = 5.4

So r=5.4

\tan {}^{ - 1}  (0)  = undefined

So our points are (5.4, undefined)

4 0
3 years ago
Which table contains a set of non-linear ordered pairs?<br><br> PLEASE HELP
Evgesh-ka [11]
It’s A because the y column contains a different set of numbers.
6 0
3 years ago
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