Answer:
Hourly
Step-by-step explanation:
if you work for that long and for that many hours you would get alot of money in about a month or two.
Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
Answer:
no solutions.
Step-by-step explanation:
8x + 14 -4x = 9+4x -5
8x-4x -4x = 9-5-14
0x = -10
anything multiplied by 0 is 0, so it's impossible to get -10 by multiplying by 0
Answer:
0.75
Step-by-step explanation:
Given,
P(A) = 0.6, P(B) = 0.4, P(C) = 0.2,
P(A ∩ B) = 0.3, P(A ∩ C) = 0.12, P(B ∩ C) = 0.1 and P(A ∩ B ∩ C) = 0.07,
Where,
A = event that the selected student has a Visa card,
B = event that the selected student has a MasterCard,
C = event that the selected student has an American Express card,
We know that,
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
= 0.6 + 0.4 + 0.2 - 0.3 - 0.12 - 0.1 + 0.07
= 0.75
Hence, the probability that the selected student has at least one of the three types of cards is 0.75.
Do it according to the order of operations
(P)E^MxD/A+S-
the answer would be -13