The largest possible last digit in the string of 2002 digits and number divisible by 19 or 31 is 9.
Given the first digit of a string of 2002 digits is 1 and the two digit number formed by consecutive digits within the string is divisible by 19 or 31.
We have to tell the last largest digit of such number.
Two digit numbers divisible by 19=19,38,57,76,95.
Two digit numbers divisible by 31=31,62,93,124
Number started with 1 =19
Last digit is 9
We have said that the number should be divisible by 19 or 31 not from both and started with 1.
Hence the largest possible last digit and number divisible by 19 or 31 in this string is 9.
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Answer:
It's a parameter.
Step-by-step explanation:
It is a parameter because you have consider the entire population, in this case the states, you have all the data of them (The areas). this would be an statistic if you cosulted a few of them instead.
Answer: 204 dollars in total
Step-by-step explanation: There are 72 people, and each serving serves six people. 72 divided by 6 =12.
12x17 equals $204
Hope this helped!
Answer:
1/3
Step-by-step explanation:
7-9=-2
2-8=-6
2/6 is 1/3 when simplified
What is the question? If you are only trying to expand the
expression then the answer would be:
1/4 (5y-3)+ 1/16 (12y+17)
(5y/4) – (3/4) + (12y/16) + (17/16)
1.25y – 0.75 + 0.75y + 1.0625
2y + 0.3125
If you are trying to find for y, then you forgot to equate
it to 0, that is:
2y + 0.3125 = 0
2y = -0.3125
<span>y = 0.15625</span>
