<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
- The lower limit is 0.5.
- The upper limit is 0.7.
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of
.
60 out of 100 scores are passing scores, hence 
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
- The lower limit is 0.5.
- The upper limit is 0.7.
A similar problem is given at brainly.com/question/16807970
Answer:
C
Step-by-step explanation:
Answer:
y = 57
Step-by-step explanation:
y α x
y = kx ; k = Constant of proportionality
81 = k* 54
k = 81/54
k = 1.5
Value of y when x = 38
y = kx
y = 1.5 * 38
y = 57
Hence, y = 57 when x = 38
4x² - 12x = 7
- 7 - 7
4x² - 12x - 7 = 0
4x² + 2x - 14x - 7 = 0
2x(2x) + 2x(1) - 7(2x) - 7(1) = 0
2x(2x + 1) - 7(2x + 1) = 0
(2x - 7)(2x + 1) = 0
2x - 7 = 0 or 2x + 1 = 0
+ 7 + 7 - 1 - 1
2x = 7 2x = -1
2 2 2 2
x = 3¹/₂ or x = ⁻¹/₂
Answer:
yes, it is.
Step-by-step explanation:
110,571 divided by 3 is 36857. So, since the answer is not a decimal or a fraction, 110,571 is divisable by 3