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mihalych1998 [28]
3 years ago
15

find rodrigos percent increase or decrease in sales each week. week 1, is 17 wwek 2, is 25 week 3, is 18 and week 4, is 27. his

boss said that he wants the increase sales by 15% each week.
Mathematics
1 answer:
sveticcg [70]3 years ago
3 0
<span>Answer: The multiplication factor of increase should be inverse of the multiplication factor of decrease.

    e.g. Say you have a number 100. You increase it by 25%. The multiplication factor is 5/4 i.e. when you multiply 100 by 5/4, you get 100*5/4 = 125. This is 25% more than 100. Now you want to decrease it by a certain % such that you get 100 back. Basically, 100*5/4 * x = 100 So x = 4/5 (inverse of 5/4) Hence, you decrease by 20% (the multiplication factor of 20% is 4/5) or Use this formula: cumulative % change = a + b + ab/100 You want the cumulative change to be 0. a + b + ab/100 = 0 If you know that you are increasing by 25% and want to find the % by which you should decrease to get the same number, 25 + b + 25b/100 = 0 5b/4 = -25 b = -20 So you need to decrease (hence you get the -ve sign) by 20%.</span>
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Flura [38]

Answer:

The inequality that you have is 5^{n}>2^{2n+1}+100,\,n>4. You can use mathematical induction as follows:

Step-by-step explanation:

For n=5 we have:

5^{5}=3125

2^{(2(5)+1)}+100=2148

Hence, we have that 5^{5}>2^{(2(5)+1)}+100.

Now suppose that the inequality holds for n=k and let's proof that the same holds for n=k+1. In fact,

5^{k+1}=5^{k}\cdot 5>(2^{2k+1}+100)\cdot 5.

Where the last inequality holds by the induction hypothesis.Then,

5^{k+1}>(2^{2k+1}+100)\cdot (4+1)

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5^{k+1}>2^{2k+3}+100\cdot 4

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When 2/3 of a number is increased by 20, the sum is then halved, the result obtained is the same as 2/3 of the number, increased
aleksklad [387]

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Step-by-step explanation:

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