Answer:
And the expected value for a vector of zeros and the covariance matrix is given by:
So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.
Step-by-step explanation:
Assuming that we have n observations from a dependent variable Y , given by
And for each observation of Y we have an independent variable X, given by
We can write a linear model on this way:
Where i a matrix for the error random variables, and for this case we can find the error ter like this:
And the expected value for a vector of zeros and the covariance matrix is given by:
So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.
Answer:
Amount = $14,024 (Approx)
Step-by-step explanation:
Given:
Amount received P = $13,500
Rate r = 7% = 0.07
Time t = 15 month = 15/12 = 1.25 = year
Discount rate = 6% = 0.06
Computation:
Maturity value = P[1+rt]
Maturity value = 13,500[1+(0.07)(1.25)]
Maturity value = $14,685.61
Time = 15 month - 6 month
Time = 9 month = 9/12 = 0.75 year
Amount = MV[1-dt]
Amount = 14,685.61[1-(0.75)(0.06)]
Amount = $14,024 (Approx)
Answer:
They are parallel
Step-by-step explanation:
They will have the same slope but never intersect
Answer:a) Probability P(exactly 2 bulbs rated 13watts)= 0.22
b) Probability(each bulb different rating)
= 0.24
Step-by-step explanation:
There are 6 13watts bulbs
8 18watts bulbs
4 23watts bulbs
Total bulbs = 18
a) Probability that all 3 bulbs are 18watts
Number of ways of pulling 3 bulbs = 18!/(3!×15!) = (6.4×10^15)/7.846×10^12) = 815 ways
Different ways of pulling 13watts bulbs out of 6 = 6!/(2!×4!)= 720/48=15ways
Different ways of pulling non 13 watts bulbs= 12!/(1!×11!) = 479,001,600/ 39,916,800 = 12
Number of ways total= 15×12=180waya
Therefore P(exactly 2 bulbs rated 13watts)= 180/815 =0.22
b) Probability P( all 3 bulbs are 1 from each rating)
Ways of pulling 3bulbs bulb each from 3 ratings are 4× 5 × 8= 192ways
Probability = 192/815 =0.24