Answer:
23
Step-by-step explanation:
Simplify the following:
(-66)/(-3) - 4 - -5
The gcd of -66 and -3 is -3, so (-66)/(-3) = (-3×22)/(-3×1) = (-3)/(-3)×22 = 22:
22 - 4 - -5
-(-5) = 5:
22 - 4 + 5
22 + 5 = 27:
27 - 4
| 2 | 7
- | | 4
| 2 | 3:
Answer: 23
Answer:
10) f'(x) = (x² + 6x - 3)/(x + 3)²
11) g'(x) = 1 - csc²x
Step-by-step explanation:
10. f(x) = x(1 - (4/(x + 3))
Expanding gives;
f(x) = x - (4x/(x + 3))
Differentiating gives;
f'(x) = 1 - 4/(x + 3) + 4x/(x + 3)²
Simplifying this gives;
f'(x) = [(x + 3)² - 4(x + 3) + 4x]/(x + 3)²
f'(x) = (x² + 6x + 9 - 4x - 12 + 4x)/(x + 3)²
f'(x) = (x² + 6x - 3)/(x + 3)²
11. g(x) = x + cot x
Rewriting this gives;
g(x) = x + (1/tan x)
We know that derivative of tan x is sec x while derivative of (1/tan x) is -csc²x
Thus;
g'(x) = 1 - csc²x
This can be written as
Differentiating this gives;
g'(x) = 1 - csc²x
Answer:
Step-by-step explanation:
Given that you drew all possible random samples of size 1,000 from the population of LSAT test takers
Plotted the values of the mean from each sample
As per central limit theorem, we find that for sample sizes of minimum 30, the means of all such samples randomly drawn would follow a bell shaped curve with normal distribution irrespective of the original distribution of the population.
Hence here answer is yes, this would be a normal distribution as per central limit theorem.
Answer:
Brian because if you do the math correctly it should be Brian
The the mean score (rounded to 2 DP) in the maths test across both classes is 63.67
What is the mean of scores?
The mean of scores is the sum of the scores for all students divided by the number of students
Class A:
mean score=39
mean score=sum of scores/number of students
39=sum of scores/11
sum of scores=39*11
sum of scores=429
Class B:
mean score=76
mean score=sum of scores/number of students
76=sum of scores/22
sum of scores=76*22
sum of scores=1672
overall mean score=(429+1672)/(11+22)
overall mean score= 63.67
Find out more about mean on:brainly.com/question/20118982
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