Answer:
Step-by-step explanation:
GIVEN: A bibliophile plans to put a total of seven books on her marble shelf. She can choose these seven books from a mixture of works from Antiquity and works on Post-modernism, of which there are seven each.
TO FIND: If the shelf must contain at least four works from Antiquity, and one on Post-Modernism, then how many ways can he select seven books to go on the shelf.
SOLUTION:
Total number of books available 
As there must be at least 
books on Antiquity and 
on Post-modernism
Therefore,
when there are 
Antiquity books and on Post-modernism.
total ways of selection 
when there are 
Antiquity books and 
on Post-modernism.
total ways of selection 
when there are Antiquity books on 
books on Post-modernism.
total ways of selection 
Total ways of selection are 
Hence the total ways of selection such that there are at least books on Antiquity and on Post-modernism.
<span><span>5 quarters 2 dimes and 7 pennies
</span>
You know that the quarters and dimes together equal to the pennies, do you divide the number of coins by half, from there you find how much money worth you have of quarters and dimes, using that you use : 0.25X+0.1Y=1.45, from there you common factor by 0.05 and you divide both sides by 5 cents and you get 5x+2y=29, from there you see that you have 5 quarters, 2 dimes and 7 pennies
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
Answer:
1.9
Step-by-step explanation:
The MAD is found by <em>finding the mean, then finding how far each number is to the mean, then finding the mean of that.</em>
The mean is: <em>(1 + 8 + 3 + 1 + 3 + 3 + 7 + 5 + 1 + 3)/10</em> = 35/10 = <em>3.5</em>
The distances of all of the numbers is <em>(absolute value of the difference between 3.5 and the number) </em>: <em>2.5, 4.5, 0.5, 2.5, 0.5, 0.5, 3.5, 1.5, 2.5, 0.5</em>
The mean of those is: <em>(2.5 + 4.5 + 0.5 + 2.5 + 0.5 + 0.5 + 3.5 + 1.5 + 2.5 + 0.5) /10</em> = 19/10 = 1.9
(2x^3)*(5x^2)+(2x^3)*(4)+(1)*(5x^2)+(1)*(4)
10x^5+8x^3+5x^2+4
at x = 1 we have one tangent line and at x = 5 we have just another tangent line.
so we have the slopes, but what about the coordinates?
well, for the first one we know x = 1 and we also know f(x), let's use f(1) to get "y", and likewise we'll do the for the second one.
![\stackrel{x=1}{f(1)}=3(1)^2-15(1)\implies f(1)=-12\qquad \qquad (\stackrel{x_1}{1}~~,~~\stackrel{y_1}{-12}) \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-12)}=\stackrel{m}{-9}(x-\stackrel{x_1}{1}) \\\\\\ y+12=-9x+9\implies y=-9x-3 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bx%3D1%7D%7Bf%281%29%7D%3D3%281%29%5E2-15%281%29%5Cimplies%20f%281%29%3D-12%5Cqquad%20%5Cqquad%20%28%5Cstackrel%7Bx_1%7D%7B1%7D~~%2C~~%5Cstackrel%7By_1%7D%7B-12%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B%28-12%29%7D%3D%5Cstackrel%7Bm%7D%7B-9%7D%28x-%5Cstackrel%7Bx_1%7D%7B1%7D%29%20%5C%5C%5C%5C%5C%5C%20y%2B12%3D-9x%2B9%5Cimplies%20y%3D-9x-3%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
