The question asks:
"Mark Atilius was expecting news from his friends with whom he agreed to reveal the great secret pyramids and spent his time at a nearby inn when he caught the attention of the Egyptian sitting beside him. He was even more surprised when he talked to him.
- You're Mark Atilius, are not you? she smiled - My name is Nefertari and I have a message for you from my grandmother. You should go right away if you want to get Pharaoh's belt you've been looking for all this time.
And he passed on the parchment he had just read.
<span> AA3 + 2 = AAA
CC6 + 6 = CBB
(AB | C) -> S57 -> E73-> S47-> E57-> S43-> W26-> S18->? </span>
Task: Find out the coordinates where Mark should come.<span> "
First, you need to solve for the position from which Mark starts.
You know:
</span><span>AA3 + 2 = AAA
Since 3 + 2 = 5,
553 + 2 = 555
Therefore A = 5.
Similarly:
</span><span>CC6 + 6 = CBB
Since 6+6 = 12, B = 2.
In order from the middle digit to be 2, the original one must have been 1.
Therefore B = 2 and C = 1
Hence, the starting position is: (AB, C) = (52, 1)
The following line gives you how many steps and in what direction Mark should go: S = south (negative vertical motion), N = north (positive vertical motion), E = east (positive horizontal motion), W = west (negative horizontal motion).
(52, 1)
-> S57 -> (52, -56)</span>
-> E73 -> (125, -56)
-> S47 -> (125, -103)
-> E57 -> (182, -103)
-> S43 -> (182, -146)
-> W26 -> (156, -146)
-> S18 -> (156, -164)
Hence, the coordinates that Mark should reach are (156, -164)
Answer:
2/3
Step-by-step explanation:
Answer:
The answer is "It would decrease, but not necessarily by 8%".
Step-by-step explanation:
They know that width of the confidence level is proportional to a confidence level. As just a result, reducing the confidence level decreases the width of a normal distribution, but not with the amount of variance in the confidence level. As just a result, when a person teaches a 90% standard deviation rather than a 98 percent normal distribution, the width of the duration narrows.
The 84th day since the greatest common multiple is 84.
