Answer:
Since the null hypothesis is true, finding the significance is a type I error.
The probability of the year I error = level of significance = 0.05.
so, the number of tests that will be incorrectly found significant is computed as follow: 0.05 * 100 = 5
Therefore, 5 tests will be incorrectly found significant given that the null hypothesis is true.
The sum of the first n natural numbers and 0=
n*(n+1)/2
44,000 *10
440,000
when multiplying by ten, add one more zero to the end of the number
The answer I believe would be 2Y + 3x=6