The amount of waste is 22 oz
<h3>How to determine the waste amount?</h3>
The weights of the 3 rainbow trouts are:
3 Ib 5 oz, 2 Ib 12 oz, and 3 Ib 8 oz.
The total weight is:
Total = 3 Ib 5 oz + 2 Ib 12 oz + 3 Ib 8 oz
Evaluate the sum
Total = 8 Ib 25 oz
The amount she was left with is 8 Ib 2 oz
Subtract this amount from the total
Waste = 8 Ib 25 oz - 8 Ib 2 oz
Evaluate
Waste = 22 oz
Hence, the amount of waste is 22 oz
Read more about sum and difference at:
brainly.com/question/10739762
#SPJ1
Answer:
something
Step-by-step explanation:
sry i need answers
Hello! So, the toy company makes blocks that cost $5 to make. Then, they mark it up by 300%, which is marking the price up by 4. 100% markup is doubling the price, 200% is tripling the price, and 300% marks the price up by 4. Then, the blocks would cost $20. After that, the price would be marked up by 150%, so that’s 2.5 times the price of the original. When you do $20 by 2.5, then the blocks cost $50. The local toy store will mark the block up by 200%, which as said before, is tripling the price. Then, you do 50 * 3, and then the price of the blocks is $150. Then, it gets marked off by 30% and 10% of 150 is 15, so 30% off 150 is 45. When you do 150 - 45, the difference is $105. Then, you take off an additional 15%, But you would still have to pay 85% of the original price. You would do 105 * 0.85, and then you would get $89.25 for the blocks. But that’s not all. You have to pay 8% sales tax for the blocks. You can find the tax by doing 89.25 * 0.08 and then adding the prices together. Or, you can do 89.25 * 1.08 to find the total price. Either way, the product should be 96.39. The total price you paid for the blocks is $96.39.
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
Note: you may have more offers of answers if you post similar questions in the computer and technology section.
