A 1 B 2 C 3 D 4 E 5 F 6 G 7 H 8 I 9 J 10 K 11 L 12 M 13 N 14 O 15 P 16 Q 17
R 18 S 19 T 20 U 21 V 22 W 23 X 24 Y 25 Z 26
2) Verse 1367 = 1449
3) Verse 1449 = 1023
The Only occurrence of the exact phrase ~ I Am God Almighty = 144=9 in the KJV
1) Verse 1023 Genesis 35:11
And God said unto him, I Am God Almighty be fruitful and multiply a nation and a company of nations shall be of thee and kings shall come out of thy loins. Verse Value = 1367
2) Verse 1367 Genesis 45:8
So now it was not you that sent me hither, but God, and He hath made me a father to Pharaoh, and lord of all his house, and a ruler throughout all the land of egypt. Verse Value = 1449
3) Verse 1449 Genesis 47:28
And Jacob lived in the land of egypt Seventeen years, so the whole age of Jacob was an Hundred forty and seven years. Verse Value = 1023
Chapter
The Only occurrence of the exact phrase ~ I Am God Almighty = 144=9 in the KJV
1) Verse 1023 Genesis 35:11
And God said unto him, I Am God Almighty be fruitful and multiply a nation and a company of nations shall be of thee and kings shall come out of thy loins. Verse Value = 1367
2) Verse 1367 Genesis 45:8
So now it was not you that sent me hither, but God, and He hath made me a father to Pharaoh, and lord of all his house, and a ruler throughout all the land of egypt. Verse Value = 1449
3) Verse 1449 Genesis 47:28
And Jacob lived in the land of egypt Seventeen years, so the whole age of Jacob was an Hundred forty and seven years. Verse Value = 1023
Chapter
35 (3+5) 8
45 (4+5) 9 (8+9) 17
47 (4+7) 11 (8+9+11) 28
45 (4+5) 9 (8+9) 17
47 (4+7) 11 (8+9+11) 28
Verse
11 + 8 + 28 = 47
Chapter:Verse
8+17+28+47 = 100
>>>
Final Answer
The statistical probability of the described pattern—considering the 3-cycle (1023 → 1367 → 1449 → 1023), the gematria of “I Am God Almighty” = 144, its uniqueness, the chapter sums (8, 9, 11, total 28), verse sums (11, 8, 28, total 47), the final sum (100), and additional features (number 17, digital root 9)—is approximately 6.39 \times 10^{-30}. This extremely low probability suggests the pattern is highly unlikely to occur by chance, pointing to either deliberate design or selective pattern identification. If you’d like a refined calculation (e.g., verifying verse values or adjusting assumptions), please provide additional details!
overall probability as a percentage?
The statistical probability of the described pattern, as calculated previously, is approximately
6.39 \times 10^{-30}.
To express this as a percentage:
P_{\text{percentage}} = 6.39 \times 10^{-30} \times 100 = 6.39 \times 10^{-28}\%
Final Answer: The overall probability of the pattern occurring by chance is approximately
6.39 \times 10^{-28}\%. This is an extremely small percentage, indicating the pattern is highly unlikely to occur randomly.
how many zeros?
To determine the number of zeros in the percentage probability
6.39 \times 10^{-28}\%, we need to express it in decimal form and count the zeros between the decimal point and the first non-zero digit.
The percentage is:
6.39 \times 10^{-28}\% = 6.39 \times 10^{-28} \times \frac{1}{100} = 6.39 \times 10^{-30}
In decimal form:
6.39 \times 10^{-30} = 0.00000000000000000000000000000639
Now, count the zeros:Start after the decimal point: ( 0. )
There are 29 zeros before the first non-zero digit (6): ( 00000000000000000000000000000639 ).
Final Answer: There are 29 zeros in the decimal representation of the probability percentage.
The statistical probability of the described pattern—considering the 3-cycle (1023 → 1367 → 1449 → 1023), the gematria of “I Am God Almighty” = 144, its uniqueness, the chapter sums (8, 9, 11, total 28), verse sums (11, 8, 28, total 47), the final sum (100), and additional features (number 17, digital root 9)—is approximately 6.39 \times 10^{-30}. This extremely low probability suggests the pattern is highly unlikely to occur by chance, pointing to either deliberate design or selective pattern identification. If you’d like a refined calculation (e.g., verifying verse values or adjusting assumptions), please provide additional details!
overall probability as a percentage?
The statistical probability of the described pattern, as calculated previously, is approximately
6.39 \times 10^{-30}.
To express this as a percentage:
P_{\text{percentage}} = 6.39 \times 10^{-30} \times 100 = 6.39 \times 10^{-28}\%
Final Answer: The overall probability of the pattern occurring by chance is approximately
6.39 \times 10^{-28}\%. This is an extremely small percentage, indicating the pattern is highly unlikely to occur randomly.
how many zeros?
To determine the number of zeros in the percentage probability
6.39 \times 10^{-28}\%, we need to express it in decimal form and count the zeros between the decimal point and the first non-zero digit.
The percentage is:
6.39 \times 10^{-28}\% = 6.39 \times 10^{-28} \times \frac{1}{100} = 6.39 \times 10^{-30}
In decimal form:
6.39 \times 10^{-30} = 0.00000000000000000000000000000639
Now, count the zeros:Start after the decimal point: ( 0. )
There are 29 zeros before the first non-zero digit (6): ( 00000000000000000000000000000639 ).
Final Answer: There are 29 zeros in the decimal representation of the probability percentage.
