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Margarita [4]
3 years ago
8

I do not understand.

Mathematics
1 answer:
Dominik [7]3 years ago
3 0
What is the question?
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En una botella caben 2.75 litros de agua. ¿Aproximadamente cuántos galones caben en la botella?1 espacio l i t r o casi igual a
Naddik [55]

Answer:

0.726 galones

Step-by-step explanation:

De la pregunta anterior, se nos dice que

1 botella = 2.75 litros de agua

Se nos pide encontrar cuántos galones puede contener.

En la pregunta, nos dicen

1 litro = 0.264 galones

2.75 litros =

Multiplicación cruzada

2.75 litros × 0.264 galones

= 0.726 galones.

Por lo tanto, 0.726 galones pueden caber en la botella

6 0
3 years ago
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
Which equation is y +5 = 1/3 (x - 9) in slope-intercept form
Alinara [238K]

Answer:

y=1/3x-8

Step-by-step explanation:

y+5=1/3(x-9)

y=1/3x-9/3-5

y=1/3x-3-5

y=1/3x-8

Please mark me as Brainliest if you're satisfied with the answer.

3 0
3 years ago
For her phone service, Lucy pays a monthly fee of $25, and she pays an additional $0.06 per minute of use. The least she has bee
Oliga [24]

Answer:

1312.5 minutes

Step-by-step explanation:

For her phone service, Lucy pays a monthly fee of $25, and she pays an additional $0.06 per minute of use. The least she has been charged in a month Is $103.72

What are the possible numbers of minutes she has used her phone in a month?

Use m for the number of minutes, and solve your inequality for m.

The term least = ≥ = greater than or equal to

Our inequality equation is given as:

$25 + $0.06 × m ≥ $103.72

25 + 0.06m ≥ 103.72

0.06m ≥ 103.72 - 25

0.06m≥ 78.75

m ≥ 78.75/0.06

m ≥ 1312.5 minutes

The possible number of minutes she used is 1312.5 minutes

6 0
3 years ago
Casey puts wrapping paper around a gift box. She wants to tie a ribbon
kkurt [141]
Well 12 inches is a foot so 24inches is 2 feet so there is a hint so 90 divided by 12 and there is part of it
3 0
3 years ago
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