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kupik [55]
2 years ago
5

The label on a large bag of flour states that it has 2 times as many cups of flour as a small bag. Casey measures 16 cups of flo

ur in the small bag.
Which multiplication sentence shows how many cups of flour are in the large bag? (A) 2×2=4
(B) 2×8=16
(C) 2×16=32
(D) 16×16=256
Mathematics
1 answer:
inna [77]2 years ago
7 0
It would be B because it has 16 cups
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s = r1 s = 3.1 s = 3.1

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v =a/s

v = 208/3.1

v = 67.0968

Step-by-step explanation:

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The pattern 10, 12, 14, 16, ________ follows the rule add two. What is the next number in the pattern?
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Solve each equation? (n+2) ( n+1)= 0
Delicious77 [7]

Answer:

n=-2,\:n=-1

Step-by-step explanation:

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7 0
3 years ago
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Three dice are rolled. Let the random variable x represent the sum of the 3 dice. By assuming that each of the 63 possible outco
tamaranim1 [39]
<h3>Answer:  1/8</h3>

In decimal form, 1/8 = 0.125 which converts to 12.5%

==================================================

Work Shown:

The 63 should be 6^3. There are 6 choices per slot, and 3 slots, so 6^3 = 216 different outcomes.

Here are all of the ways to add to 11 if we had 3 dice

  1. sum = 1+4+6 = 11
  2. sum = 1+5+5 = 11
  3. sum = 1+6+4 = 11
  4. sum = 2+3+6 = 11
  5. sum = 2+4+5 = 11
  6. sum = 2+5+4 = 11
  7. sum = 2+6+3 = 11
  8. sum = 3+2+6 = 11
  9. sum = 3+3+5 = 11
  10. sum = 3+4+4 = 11
  11. sum = 3+5+3 = 11
  12. sum = 3+6+2 = 11
  13. sum = 4+1+6 = 11
  14. sum = 4+2+5 = 11
  15. sum = 4+3+4 = 11
  16. sum = 4+4+3 = 11
  17. sum = 4+5+2 = 11
  18. sum = 4+6+1 = 11
  19. sum = 5+1+5 = 11
  20. sum = 5+2+4 = 11
  21. sum = 5+3+3 = 11
  22. sum = 5+4+2 = 11
  23. sum = 5+5+1 = 11
  24. sum = 6+1+4 = 11
  25. sum = 6+2+3 = 11
  26. sum = 6+3+2 = 11
  27. sum = 6+4+1 = 11

There are 27 ways to add to 11  using 3 dice. This is out of 216 total outcomes of 3 dice being rolled.

So, 27/216 = (1*27)/(8*27) = 1/8 is the probability of getting 3 dice to add to 11.

7 0
2 years ago
HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x
Anna35 [415]

 

\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0


\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}




3 0
3 years ago
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