srikanth has recently got married and he wants to maintain his monthly expenses. He recorded his monthly expenses E ( n ) for last year as E ( n ) = − n 2 + 16 n − 40 , where n is the number of month and n ∈ { 1 , 2 , . . . , 11 , 12 } . He came to know that he should keep his monthly expense ₹20,000 per month for fulfilling his other plans. Let N 1 be the set of month numbers when he spent more than ₹20,000, N − 1 be the set of month numbers when he spent less than ₹20,000, and N 0 be the set of month numbers when he spent ₹20,000. Assuming last year expenses as the benchmark for his expenses, choose the set of correct options. Cardinality of N 1 is 3. Cardinality of N − 1 is 7. Based on the cardinalities of N 1 and N − 1 only, we can not prove that Srikanth has spent on average ₹20,000 per month last year. As cardinality of N 1 is more than N − 1 , gives a proof that Srikanth has spent on average less than ₹20,000 per month last year. As cardinality of N 1 is equal to N − 1 , gives a proof that Srikanth has spent on average ₹20,000 per month last year.
Answers
Answer:
Cardinality of N 1 is 3.
Cardinality of N -1 is 7.
Step-by-step explanation:
Given : E(n) = -n² + 16n - 40
E(n) in thousands
n is the number of month and n∈{1,2,...,11,12}.
should keep his monthly expense ₹20,000 per month for fulfilling his other plans
To Find : Cardinality of N₁ , N₋₁ and N₀
Solution:
E(n) = -n² + 16n - 40
E(n) = 20 thousands
-n² + 16n - 40 = 20
=> -n² + 16n - 60 = 0
=> -( n² - 16n + 60) = 0
=> -(n - 10)(n - 6) = 0
n = 6 , n = 10
Expense = 20 k
n< 6 and n > 10 Expense is less
6 < n < 10 Expense is more
N₁ = { 7 , 8 , 9 } Cardinality of N₁ = 3
N₋₁ = { 1 , 2 ,3 , 4 , 5 , 11 , 12} Cardinality of N₋₁ = 7
N₀ = { 6 , 10} Cardinality of N₀ = 2
As Cardinality of N₋₁ is more than Cardinality of N₁
Hence he spend less than 20k on average
Learn More
write the set of even prime number in rule method - Brainly.in
brainly.in/question/6778533
find the truth value of all prime numbers are either even or odd ...
brainly.in/question/9141870