Answer:
Let the scores on the first two test be x
Let score for third test be 14 exceeds second( or first) = x + 14
Total = 233
x + x + (x + 14) = 233
3x + 14 = 233
3x = 219
x = 73
Score of the third test = x + 14 = 73 + 14 = 87
Therefore, the scores are 73, 73, 87
Answer:
yes 2=138 degrees
Step-by-step explanation:
Not hundo perecent sure, sorry if it's wrong, if you want me to show my work then please ask
Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.
Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.
Let us try it below:
Sigma notation 1:
10
<span> Σ (2i + 3)
</span>i = 3
@ i = 3
2(3) + 3
12
The first sigma notation does not have the same result, so we move on to the next.
10
<span> Σ (3i + 2)
</span><span>i = 3
</span>
When i = 3; <span>3(3) + 2 = 11. (OK)
</span>
Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.
When i = 4; <span>3(4) + 2 = 14. (OK)
</span>When i = 5; <span>3(5) + 2 = 17. (OK)
</span>When i = 6; <span>3(6) + 2 = 20. (OK)
</span>When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; <span>3(8) + 2 = 26. (OK)
</span>When i = 9; <span>3(9) + 2 = 29. (OK)
</span>When i = 10; <span>3(10) + 2 = 32. (OK)
Adding the results from her 3rd through 10th trials: </span><span>11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.
</span>
Therefore, the total jumps she had made from her third to tenth trips is 172.
Answer: 6.19
Step-by-step explanation: 106.25 divided by 17 is around 6.19.
Answer:
44.88
Step-by-step explanation:
Heaviest box = 424.25
Lightest box = 379.37
424.25 - 379.37 = 44.88
Hope this helps!
