Answer:
This sample is not representative of all UF students, since those who are not local are not considered.
Step-by-step explanation:
This is a common statistics practice, when we want to study something from a population, we find a sample of this population.
However, the sample has to be representative
For example:
I want to estimate the proportion of New York state residents who are Buffalo Bills fans. So i ask, lets say, 1000 randomly selected Buffalo residents wheter they are Buffalo Bills fans, and expand this to the entire population of New York State residents. This is not representative of all New York State residents, just Buffalo residents.
In this problem, we have that:
They conduct a phone survey (with local numbers selected at random from the student directory) calling people during "dead week". Will this sample be representative of all UF students?
They only call those students with local numbers.
However, in an university, it is expected that there will be a good percentage of non local students.
So this sample is not representative of all UF students, since those who are not local are not considered.
Answer:4
Step-by-step explanation:i cant rlly explain
In Δ ABC, ∠A=120°, AB=AC=1
To draw a circumscribed circle Draw perpendicular bisectors of any of two sides.The point where these bisectors meet is the center of the circle.Mark the center as O.
Then join OA, OB, and OC.
Taking any one OA,OB and OC as radius draw the circumcircle.
Now, from O Draw OM⊥AB and ON⊥AC.
As chord AB and AC are equal,So OM and ON will also be equal.
The reason being that equal chords are equidistant from the center.
AM=MB=1/2 and AN=NC=1/2 [ perpendicular from the center to the chord bisects the chord.]
In Δ OMA and ΔONA
OM=ON [proved above]
OA is common.
MA=NA=1/2 [proved above]
ΔOMA≅ ONA [SSS]
∴ ∠OAN =∠OAM=60° [ CPCT]
In Δ OAN


OA=1
∴ OA=OB=OC=1, which is the radius of given Circumscribed circle.
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content…
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side.
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day.
Answer:
I'm not sure but I think the answer is >
Step-by-step explanation:
It looks rig but if I were u I wouldn't trust me