20,0000000000000000000000 I can’t help with this I dropped out of school
Based on the data recorded by Isabella, it can be concluded the cube is rather fair.
<h3>How many times did Isabella get each number?</h3>
Based on the data, here are the results:
- Getting a 1: 6 times
- Getting a 2: 5 times
- Getting a 3: 7 times
- Getting a 4: 5 times
- Getting a 5: 6 times
- Getting a 6: 7 times
This implies, in total Isabella got the same number between five and seven times. For example, the number 2 was obtained 5 times, but the number 3 was obtained 7 times.
<h3>What can be concluded based on the results?</h3>
Even though Isabella did not get the same number of times each number, the dice is rather fair because by rolling the dice thirty six times you will obtain the same number at least five times.
Moreover, there is not a big difference in the number of times you obtain each number.
Learn more about dice in: brainly.com/question/23637540
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Well you didn't post the expressions but let X be the amount of money for first place. This means X-50 would be the amount of second place because <span>each subsequent place after 1st wins $50 less than the previous place. Thirds, would be X-50-50 or X-100
Thus the total amount of money would be all of the values combined which is X+X-50+X-100=3X-150
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Answer: the number, converted to base 10 is 884
Step-by-step explanation:
The table gives symbols for a base 12 numerical system and it is called Caidoz numbers.
Looking at the table, the corresponding Caidozian number that we considering is 618 to base 12. Converting 618 to base 10, it becomes
8 × 12^0 = 8 × 1 = 8
1 × 12^1 = 1 × 12 = 12
6 × 12^2 = 6 × 144 = 864
Therefore, 618 to base 12 to 618 to base 10 would be
8 + 12 + 864 = 884
Answer:
• Group the h terms by organised term arrangement :
• Then using distributive property, factorise out the value h so that the reverse is true.
• for the variable "lw", divide it by h in order to add it to the bracket of (w + l). Make sure the reverse is true:
• finally, completely factorise out the value h
