Step-by-step explanation:
Aight, so the same intercept

m=½

soooo

Answer: 50 texts
Step-by-step explanation: 20.8 - 5.80 = 15
15 ÷ .30 = 50
Draw the graph of R = {(1,1), (2,2), (3,3), (4,4), (5,5), (1,4), (2,5), (4,1), (5,2)}
tresset_1 [31]
Answer:
Step-by-step explanation:
(-4, 6), (0, 2), (4, -2) are all ordered pairs that have the sum as 2.
-4+6 is the same thing as 6-4 if there's more positive than negative
0+2=2 because 0 plus a number equals to that number
4-2=2 because you're adding a negative number with the positive number together to get the answer
Just be aware that I might not make any sense at all
Answer:
a) For the 90% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
b) For the 99% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
Step-by-step explanation:
Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
Solution to the problem
Part a
For the 90% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
Part b
For the 99% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got: