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DeathFightervx
23.07.2020 •
Mathematics
The ability to find a job after graduation is very important to GSU students as it is to the students at most colleges and universities. Suppose we take a poll (random sample) of 3957 students classified as Juniors and find that 3226 of them believe that they will find a job immediately after graduation. What is the 99 % confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation?
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Ответ:
Step-by-step explanation:
Using the formula
p +/- z* √(p(1-p) / n)
Where p is sample proportion = 3226/3957 = 0.8153, n = 3957 and z* is 2.576
0.8153 + (2.576 √(0.8153(1-0.8153) / 3957))
0.8153 + (2.576 √ (0.8153(0.1847) /3957)
0.8513 + (2.576 √(0.1572/3957)
0.8513 + (2.576 √0.00003974)
0.8513 + 2.576(0.0063)
0.8513 + 0.016
0.8670 ~ 87%
For the lower interview
0.8513 - 0.016
= 0.8353 ~ 84%
Thus, at 99% confidence interval is 84% and 87%
Ответ:
The correct answer is D. x = 11.1
Step-by-step explanation:
The two chords in the circle are intersecting with each other and are divided into two sub parts.
By Chord intersection theorem we have : If the two chords intersect with each other then the product of their corresponding sub parts is equal.
So, to find x :
⇒ 26 × 6 = x × 14
⇒ 14·x = 156
⇒ x = 11.1 units
Therefore, The correct answer is D. x = 11.1