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balu736 [363]
1 year ago
9

A pound of popcorn is popped for a class party. The popped corn is put into small popcorn boxes that each hold 120 popped kernel

s. There are 1,600 kernels in a pound of unpopped popcorn. If all the boxes are filled except for the last box, how many boxes are needed and how many popped kernels are in the last partially filled box?
Mathematics
2 answers:
Aleksandr [31]1 year ago
5 0
13 of the boxes would be entirely filled, so 14 boxes would be needed in total. There would be 40 popped kernels in the last box.
Phantasy [73]1 year ago
3 0
1600 / 120 = 13.3
You will need 14 boxes.

120 x 0.3 = 36
There were 36 kernels in the partially filled box.
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Jack works as a waiter and is keeping track of the tips he earns daily. About how much does Jack have to earn in tips on
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25$

Step-by-step explanation:

Add all of the days up and divide by 7

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Use synthetic division to solve (x^3 + 1) ÷ (x – 1). What is the quotient?
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A cube with side length 4p is stacked on another cube with side length 2q^2. What is the total volume of the cubes in factored f
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Volume of cube=side³
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1 year ago
ian has a bag of fruit chews. 60/100 of fruit chews are cherry or grape. if 25/100 of fruit chews are cherry what fraction of th
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1 year ago
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Using the extended Euclidean algorithm, find the multiplicative inverse of a. 135 mod 61 b. 7465 mod 2464 c. 42828 mod 6407
rodikova [14]

Answer:

(a)1≡47 mod 61

(b)1≡2329 mod 2464

(c)Does not exist

Step-by-step explanation:

The operation a(mod b) has an inverse if the the two integers (a,b)

are co-prime. i.e. their g.c.d is 1.

(a)Given 135 mod 61

We first reduce it to its lowest form.

135 mod 61=13 mod 61

61=13(4)+9 ==> 9=61-13(4)

13=9(1)+4 ==> 4=13-9(1)

9=4(2)+1 ==> 1=9-4(2)

4=1(4)

Next we rewrite 1 as a linear combination of 13 and 61.

1=9-4(2)

=9-(13-9(1))2

=9(3)-13(2)

=(61-13(4))(3)-13(2)

=61(3)-13(12)-13(2)

1=61(3)-13(14)

1=61(3)+13(-14)

1≡-14 mod 61≡(-14+61)mod 61

1≡47 mod 61

(b)7465 mod 2464

Reducing it to its lowest form

7465 mod 2464=73 mod 2464

2464=73(33)+55 ==>55=2464-73(33)

73= 55(1)+18 ==> 18=73-55(1)

55=18(3)+1 ==>1=55-18(3)

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Rewriting 1 as a linear combination of 73 and 2464.

1=55-18(3)

=2464-73(33)-(73-55(1))(3)

=2464-73(33)-73(3)+55(3)

=2464-73(36)+55(3)

=2464-73(36)+(2464-73(33))(3)

=2464-73(36)+2464(3)-73(99)

=2464(4)-73(135)

1=2464(4)+73(-135)

Therefore:

1≡-135 mod 2464

1≡(-135+2464)mod 2464

1≡2329 mod 2464

(c)42828 mod 6407

The two numbers are not co-prime. In fact their g.c.d is 43.

Therefore their inverse does not exist.

4 0
2 years ago
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