Answer:
4x
Step-by-step explanation:
Answer:
- 9
Step-by-step explanation:
The average rate of change of f(x) in the closed interval [ a, b ] is
Here [a, b ] = [ - 5, 2 ], thus
f(b) = f(2) = 2² - 6(2) + 2 = 4 - 12 + 2 = - 6
f(a) = f(- 5) = (- 5)² - 6(- 5) + 2 = 25 + 30 + 2 = 57 , thus
average rate of change = 
= 
= - 9
Answer:
400 units³
Step-by-step explanation:
The volume (V) of the square pyramid is
V = 
area of base × height (h)
where h is the perpendicular height.
Consider the right triangle formed by a segment from the vertex to the midpoint of the base and the slant height ( the hypotenuse )
Using Pythagoras' identity on the right triangle
h² + 5² = 13²
h² + 25 = 169 ( subtract 25 from both sides )
h² = 144 ( take the square root of both sides )
h = 
= 12
Area of square base = 10² = 100, hence
V = × 100 × 12 = 4 × 100 = 400
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
_____
My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
Answer:
the answer is slope be it give U just the formula
