600000+90000+2000+4=692004
I think that's how you write it.
Answer: 0.7
Step-by-step explanation:
p(tail) + p(head) = 1
p(tail) = 1 - p(head)
p(tail) = 1 -0.3
p(tail) = 0.7
mark the brainliest plzz
Answer:
The set of natural numbers can be denoted as: {} (null set) because no natural number will satisfy the given equation.
Step-by-step explanation:
Given equation:
To find the set of natural numbers that satisfies the above equation.
Solution:
In order to find the set of solutions for the equation, we will solve it for 
We have:
Subtracting both sides by 8.
∴ 
Thus, we get only one solution for that satisfies the given equation but it is not a natural number. -3 is an integer.
Set of natural numbers {1,2,3,4,5,6........}
Thus, we can say the set of natural numbers that satisfies the equation is a null set {}.
These are two questions and two answers.
1) Problem 17.
(i) Determine whether T is continuous at 6061.
For that you have to compute the value of T at 6061 and the lateral limits of T when x approaches 6061.
a) T(x) = 0.10x if 0 < x ≤ 6061
T (6061) = 0.10(6061) = 606.1
b) limit of Tx when x → 6061.
By the left the limit is the same value of T(x) calculated above.
By the right the limit is calculated using the definition of the function for the next stage: T(x) = 606.10 + 0.18 (x - 6061)
⇒ Limit of T(x) when x → 6061 from the right = 606.10 + 0.18 (6061 - 6061) = 606.10
Since both limits and the value of the function are the same, T is continuous at 6061.
(ii) Determine whether T is continuous at 32,473.
Same procedure.
a) Value at 32,473
T(32,473) = 606.10 + 0.18 (32,473 - 6061) = 5,360.26
b) Limit of T(x) when x → 32,473 from the right
Limit = 5360.26 + 0.26(x - 32,473) = 5360.26
Again, since the two limits and the value of the function have the same value the function is continuos at the x = 32,473.
(iii) If T had discontinuities, a tax payer that earns an amount very close to the discontinuity can easily approach its incomes to take andvantage of the part that results in lower tax.
2) Problem 18.
a) Statement Sk
You just need to replace n for k:
Sk = 1 + 4 + 7 + ... (3k - 2) = k(3k - 1) / 2
b) Statement S (k+1)
Replace
S(k+1) = 1 + 4 + 7 + ... (3k - 2) + [ 3 (k + 1) - 2 ] = (k+1) [ 3(k+1) - 1] / 2
Simplification:
1 + 4 + 7 + ... + 3k - 2+ 3k + 3 - 2] = (k + 1) (3k + 3 - 1)/2
k(3k - 1)/ 2 + (3k + 1) = (k + 1)(3k+2) / 2
Do the operations on the left side and you will find it can be simplified to k ( 3k +1) (3 k + 2) / 2.
With that you find that the left side equals the right side which is a proof of the validity of the statement by induction.
D ope this helped please leave a thank you
