Answer:
369.7 mL of medication
Step-by-step explanation:
How many mL of medication are needed to last 10 days if the dose of medication is 2.5 tsp TID (three times a day)?
From the above question,
The dosage of the medication =
2.5 tsp 3 times a day
= 2.5 × 3 = 7.5 tsp per day.
Since
1 day = 7.5 tsp
10 days = x tsp
Cross Multiply
x = 10 × 7.5 tsp
x = 75 tsp of medication for 10 days.
Step 2
It is important to note that:
1 tsp = 4.929 mL
75 tsp = x mL
Cross Multiply
x = 75 × 4.929 mL
x = 369.669 mL of medication
Approximately = 369.7 mL of medication
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.
Hi there
The formula is
A=p (1+r/k)^kt
A future value 3000
P present value 100
R interest rate 0.02
K compounded monthly 12
T time?
We need to solve for t
T=[log (A/p)÷log (1+r/k)]÷k
T=(log(3,000÷100)÷log(1+0.02÷12))÷12
T=170.202 years
So it's a
Hope it helps
Solution:
<u>Note that:</u>
- Given inequality: 12p < 96
<u>Dividing both sides by 12:</u>
- 12p < 96
- => 12p/12 < 96/12
- => p < 8
Correct option is A.
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