A. The number of 10-boards Peter bought is equal to n divided by 10. Then, each of the 10-boxes will get two boxes of nails. The number of boxes of nails that Peter will have after buying n boards will be,
N = (2)(n/10)
Simplifying,
<em> N = n/5</em>
b. If the number of boards are 90 then,
N2 = (90/10)(2)(100 nails/box)
N2 = 1800
Answer: 1800
Answer:
B is the answer.
Step-by-step explanation:
![x^{\frac{m}{n} } = \sqrt[n]{x^{m} }](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%20%7D%20%20%3D%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%20%7D)
![\sqrt[4]{81^{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B81%5E%7B5%7D%20%7D)
= 
Perhaps the easiest way to solve this problem is to convert 13 into a fraction that has the same denominator as 7 5/8.
Convert both to improper fractions:
7 5/8 turns into 61/8, and
13 turns into 104/8.
Then, subtract 61/8 from 104/8:
104/8-61/8=43/8.
Simplify (mixed fraction):
5 3/8.
The second ribbon has a length of 5 3/8 meters.
Answer:
3285
Step-by-step explanation:
5x3144=15720
3990
15720+3990=19710
19710/6=3285
11, 12, 15, 16, 20, 20, 25
Mean: 17
The range is 25-11 = 14 (the numbers in bold)
The interquartile range is 20-12=8 (the numbers underlined)
The mean absolute deviation is: 4; this is found by finding how far each number is from 17 (mean): 6,5,2,1,3,3,8 (28) and dividing by 7.
Part B:
The prices vary by no more that $14 (range).
<span>The middle half of the prices vary by no more than $8 (IQR).
</span>
<span>The admission prices differ from the mean price by an average of $4 (MAD).</span>
