Answer:
Point A is the center of the circle that passes through points E, F, and G and the center of the circle that passes through points X, Y, and Z.
Step-by-step explanation:
A is the intersection of angle bisectors, so is the incenter of triangle EFG. It is also the intersection of the perpendicular bisectors of the sides of triangle EFG, so is the circumcenter.
The altitudes at X, Y, and Z are perpendicular to sides EF, EG, and FG, and pass through the incenter, so X, Y, Z are points on the incircle.
A is the center of circles through E, F, and G, and through X, Y, and Z.
Answer:
28a 20b 24
Step-by-step explanation:
Answer:
30°
Step-by-step explanation:
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:
Answer:
63°
Step-by-step explanation:
TSU = 180 - 117 = 63