Answer:
The sequence is:
5, 14, 23, 32
Step-by-step explanation:
We know that an Arithmetic sequence with a₁ and the commo difference 'd' has the nth term such as:
aₙ = a₁ + (n-1)d
Given the arithmetic sequence
5, _____, ______, 32
here:
a₁ = 5
n = 4
aₙ
so substituting a₁ = 5, n = 4 and aₙ = 32 in the nth term
aₙ = a₁ + (n-1)d
32 = 5 + (4-1)d
32 = 5 + 3d
3d = 32-5
3d = 27
divide both sides by 3
3d/3 = 27/3
d = 9
Therefore, the common difference: d = 9
Determining the 2nd term:
Using the formula
aₙ = a₁ + (n-1)d
substitute n = 2, a₁ = 5, d = 9
a₂ = 5 + (2-1)9
a₂ = 5 + 1(9)
a₂ = 5 + 9
a₂ = 14
Determining the 3rd term:
Using the formula
aₙ = a₁ + (n-1)d
substitute n = 3, a₁ = 5, d = 9
a₃ = 5 + (3-1)9
a₃ = 5 + 2(9)
a₃ = 5 + 18
a₃ = 23
Thus, the sequence becomes:
5, 14, 23, 32
Answer:
~ 15 %
Step-by-step explanation:
116.4 - 98.3 = 18.1
18.1 / 17.5 = 1.034 SD ABOVE the mean
this is a z-score of .8485 meaning 84.85 % have a score LESS than 116.4 so 15.15 % have a score higher
Ok first you divide -1/2 by -8 which equals your answer
It is
12000 x (1.06)^12 + 50000 x (1.06)^6
= 95,072.31
start of 4th year to end of 6th year = 6 semi-annual periods where interest is compounded for the second deposit
x^3-1
Step-by-step explanation:
because a cubic polynomial would have a cubic exponent to make it a cubic polynomial
