The boundaries of the 99% confidence interval are
Lower bound = 0.7232Upper bound = 0.7348Constructing the 99% confidence intervalFrom the question, we have the following parameters that can be used in our computation:
Selected, x = 28021
Sample, n = 38433
This means that the proportion, p is
p = 28021/38433
p = 0.729
The 99% confidence interval is then calculated as
CI = p ± z * √(p * (1 - p)/n)
Where
z = 2.576 i.e. the z-score at 99% confidence interval
So, we have
CI = 0.729 ± 2.576 * √(0.729 * (1 - 0.729)/38433)
Evaluate
CI = 0.729 ± 0.0058
Evaluate
CI = (0.7232, 0.7348)
Hence, the confidence interval is (0.7232, 0.7348)
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0.005627 to 3 decimal places
(i) Correcting the figures to 3 decimal places:
-0.005627 ≈ -0.006
0.0056 ≈ 0.006
-0.0049327 ≈ -0.005
0.0049 ≈ 0.005
-0.001342 ≈ -0.001
(ii) Correcting the figures to 3 significant figures:
-0.005627 ≈ -0.00563
0.0056 ≈ 0.00560
-0.0049327 ≈ -0.00493
0.0049 ≈ 0.00490
-0.001342 ≈ -0.00134
(i) When rounding to 3 decimal places, we look at the fourth decimal place and round the figure accordingly. If the fourth decimal place is 5 or above, we round up the preceding third decimal place. If the fourth decimal place is less than 5, we simply drop it.
(ii) When rounding to 3 significant figures, we consider the digit in the third significant figure. If the digit in the fourth significant figure is 5 or above, we round up the preceding third significant figure. If the digit in the fourth significant figure is less than 5, we simply drop it.
Rounding to the correct number of decimal places or significant figures is important to maintain precision and accuracy in calculations and measurements. It helps to ensure that the reported values are appropriate for the level of precision required in a given context.
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A small college has 204 student athletes. The number of students who play soccer is 52. The number of students who play volleyball is 31. The probability that a student plays in both volleyball and soccer is 5/204.What is the probability that a randomly selected student athlete in this school: Plays both soccer and volleyball? Plays volleyball?
To calculate the probabilities, we can use the following information:
Total number of student athletes = 204
Number of students who play soccer = 52
Number of students who play volleyball = 31
Probability of a student playing both soccer and volleyball = 5/204
1. Probability that a student plays both soccer and volleyball:
Let's denote the probability of playing both soccer and volleyball as P(Soccer and Volleyball). From the given information, we know that the number of students who play both soccer and volleyball is 5.
P(Soccer and Volleyball) = Number of students who play both soccer and volleyball / Total number of student athletes
P(Soccer and Volleyball) = 5 / 204
2. Probability that a student plays volleyball:
We want to find the probability of a student playing volleyball, denoted as P(Volleyball).
P(Volleyball) = Number of students who play volleyball / Total number of student athletes
P(Volleyball) = 31 / 204
Therefore, the probability that a randomly selected student athlete in this school plays both soccer and volleyball is 5/204, and the probability that they play volleyball is 31/204.
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5 In a Survery of 130 people 80 claimed to be CDO partisans and 60 claimed to be Anc partisan. If 30 of them are both ANC and CDO how many people are none of these two parties
Answer: there are 20 people who claimed to be neither CDO partisans nor ANC partisans.
Step-by-step explanation:
To determine the number of people who are none of these two parties, we need to subtract the total number of people who claimed to be CDO partisans, ANC partisans, and those who claimed to be both from the total number of people surveyed.
Total surveyed people = 130
Number claiming to be CDO partisans = 80
Number claiming to be ANC partisans = 60
Number claiming to be both ANC and CDO = 30
To find the number of people who are none of these two parties, we can calculate it as follows:
None of these two parties = Total surveyed people - (CDO partisans + ANC partisans - Both ANC and CDO)
None of these two parties = 130 - (80 + 60 - 30)
None of these two parties = 130 - 110
None of these two parties = 20
Triangle ABC has the following coordinates: A=(5,-5), B=(3,-3), C=(5,-3) What are the coordinates of triangle A'B'C' if it is created by dilating triangle ABC with the origin (0,0) as the center of dilation and with a scale factor of 3?
Answer:A' = (15, -15), B' = (9, -9), and C' = (15, -9)
Step-by-step explanation:
To dilate triangle ABC with a center of dilation at the origin (0,0) and a scale factor of 3, you need to multiply the coordinates of each vertex by the scale factor.
Let's calculate the coordinates of triangle A'B'C':
For point A:
x-coordinate of A' = scale factor * x-coordinate of A = 3 * 5 = 15
y-coordinate of A' = scale factor * y-coordinate of A = 3 * (-5) = -15
Therefore, A' = (15, -15)
For point B:
x-coordinate of B' = scale factor * x-coordinate of B = 3 * 3 = 9
y-coordinate of B' = scale factor * y-coordinate of B = 3 * (-3) = -9
Therefore, B' = (9, -9)
For point C:
x-coordinate of C' = scale factor * x-coordinate of C = 3 * 5 = 15
y-coordinate of C' = scale factor * y-coordinate of C = 3 * (-3) = -9
Therefore, C' = (15, -9)
Hence, the correct coordinates of triangle A'B'C' are A' = (15, -15), B' = (9, -9), and C' = (15, -9).