Answer: -3
Step-by-step explanation: If you're giving your friend 3 cookies, it's a decrease which means that it's negative.
So giving your friend 3 cookies can be written as -3.
In one year
1 + (23/100) = (1 + x)^12
1 + 0.23 = (1 + x)^12
(1 + x)^12 = 1.23
1 + x = 1.23^(1/12)
x = 1.23^(1/12) - 1
x =~ 0.017400841772181508280113627030242
That's about 1.74% per month (which is about 20.881% per year), compounded monthly, to equal 23% per year, compounded annually.
So I believe you can write
<span>I=(1+.0174<span>)<span>12t</span></span></span>or<span><span>I=<span>1.0174<span>12t</span></span></span></span>
Answer: 8
Step-by-step explanation:
<u>64</u> ÷[4* <u>27</u> (-5^2
-5 times -5 =25
27-25=2
4*2=8
64 divided by 8= 8
8 is your answer
5.8 x 2.19
= (5.00 + 0.80) x (2.00 + 0.19)
= 5.00*2.00 + 5.00*0.19 + 0.80*2.00 + 0.80*0.19
= 10.00 + 0.95 + 1.60 + 8/10 * 19/100
= 10.00 + 0.95 + 1.60 + (8*(10+9))/1000
= 10.00 + 0.95 + 1.60 + (80+72)/1000
= 10.00 + 0.95 + 1.60 + 152/1000
= 10.000 + 0.950 + 1.600 + 0.152
= 11.600 + 0.950 + 0.152
= 12.550 + 0.152
= 12.702
Hope this helps!
Answer:
0.333
Step-by-step explanation:
The first sequence starts at 1 and has a common difference of 2. It will include every odd number in the range.
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The second sequence starts at 1 and has a common difference of 3. It will include odd numbers and the even numbers 4, 10, 16, .... That is, all even numbers of the form 6n -2 will be included. The last one corresponds to the largest value of n such that ...
6n -2 ≤ 1000
6n ≤ 1002
n ≤ 167
That is, 167 even numbers will also be excluded.
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The third sequence starts at 1 and has a common difference of 4. Every number in this sequence is also a number in the first sequence.
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So the numbers in these sequences include all 500 odd numbers and 167 even numbers, for a total of 667 numbers. The probability that a randomly chosen number is not in one of these sequences is ...
(1000 -667)/(1000) = 333/1000 = 0.333 . . . . p(not a sequence term)
