Answer:
x=1
Step-by-step explanation:
5x + 2 = 3x + 4(2x - 1)
Distribute the 4
5x + 2 = 3x + 8x - 4
Combine like terms
5x + 2 = 11x - 4
Subtract 5x from each side
5x -5x + 2 = 11x - 5x - 4
2 = 6x-4
Add 4 to each side
2+4 = 6x-4+4
6 = 6x
Divide each side by 6
6/6 =6x/6
1= x
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.
By drawing?
If so, here: first outline the current shape as big as you want it... Then, erase the first shape!
Answer:
Pete
Step-by-step explanation:
Given that:
Mandy's Estimate :
Number of spins , n = 20
Pete's Estimate:
Number of spins, n = 200
A good probability estimate is one which has narrow margin of error with a high degree of confidence. These two variables are affected by sample size.
A high sample size give a narrower margin of error and increases the confidence level probability
Based on the sample size used by each of Pete and Mandy, we can conclude that, Pete's probability estimate would be better due to its significantly higher sample size.
The answer to the question is no solution.
