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AlekseyPX
2 years ago
10

You have 10 gallons of lemonade to sell. (1 gal $\approx$ 3785 cm3)

Mathematics
1 answer:
Zielflug [23.3K]2 years ago
4 0

Answer:

Step-by-step explanation:

volume of cone cup = ⅓πr²h = ⅓π4²·11 ≅ 184.31 cm³

10 gal × 3,785 cm³/gal = 37,850 cm³

37,850 cm³ × (1 cone)/(184.31 cm³) ≅205.4 cones

Buy 5 packages (250 cones).

8 gal × 3,785 cm³/gal = 30,280 cm³

30,280 cm³ × (1 cone)/(184.31 cm³) ≅165 cones

250 - 165 = 85 cones left over

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2 years ago
The cost of Nate's dinner including a 15% tip was $43.70. What was the cost of dinner alone?
Nookie1986 [14]

Answer:

The cost of dinner alone was $37.145

Step-by-step explanation:

15%=0.15----Convert your percentage into a decimal

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$43.70-$6.555=$37.145----Subtract the last number(6.555) from the total(43.70) to get the cost of dinner alone of $37.145 -This is making you subtract the percentage from the total!

I hope this helped!

7 0
2 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

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  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

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  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

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Answer:

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Just add the cost of everything

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