Answer:
b) 6.68%
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The mean score on the scale is 50. The distribution has a standard deviation of 10.
This means that 
Matthew scores a 65. What percentage of people could be expected to score the same as Matthew or higher on this scale?
The proportion is 1 subtracted by the p-value of Z when X = 65. So
has a p-value of 0.9332.
1 - 0.9332 = 0.0668
0.0668*100% = 6.68%
So the correct answer is given by option b.
Answer:
38.1 cm²
Step-by-step explanation:
area of full circle = π6² = 113.04 cm²
area of 150° segment = (150/360)(113.04) = 47.1 cm²
triangle base = sin 75° x 6 x 2 = 11.59
triangle height = cos 75° x 6 = 1.553
area of triangle = 1/2(11.59)(1.553) = 9 cm²
150° segment - triangle = 47.1 - 9 = 38.1 cm²
2876 Rounds To 3000
513 Rounds To 500
18 Rounds To 20
Then multiply:
3000x500x20= 30,000,000
Hope i helped :)
Answer:
The correct option is c which is if this test was one-tailed instead of two-tailed, you would reject the null.
Step-by-step explanation:
a: This statement cannot be true as the p-value for a 1 tailed test is dependent on the level of significance and other features.
b: This statement cannot be true as there is no valid mathematical correlation between the p-value of the one-tailed test and the current p-value.
c: This statement is true because due to the enhanced level of significance, the null hypothesis will not be rejected.
d: This statement is inverse of statement c which cannot be true.
e: The statement cannot be true as there is no correlation between the current p-value and the p-value of 1 tailed test. The correlation exists between the values of one-tailed and two-tailed p-values.
(174,000, 134,000, 420,000, 452,000, 474,000)
I don't know if i'm right for sure but I really hope this helps you! :)
