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cupoosta [38]
3 years ago
8

A bag contains 12 pieces of candy. In how many weighs can 4 pieces be selected

Mathematics
1 answer:
igomit [66]3 years ago
3 0

Answer:

495

Step-by-step explanation:

From 12, choose 4:

₁₂C₄ = 12! / (4! (12−4)!)

₁₂C₄ = 12! / (4! 8!)

₁₂C₄ = 495

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Can anybody help me with this?
mihalych1998 [28]

Answer:

Cost = 17.08

Step-by-step explanation:

The cost is 4.88 per pound.

We buy 3.5 pounds

Cost = number of pounds * cost per pound

Cost = 3.5 * 4.88

Cost = 17.08

6 0
3 years ago
three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal
Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

=     x + x + (x + \frac{6}{5}) = \frac{27}{4}

=     3x + \frac{6}{5} = \frac{27}{4}

=     3x = \frac{27}{4} - \frac{6}{5}

( let's take the LCM of 4 and 5 = 20

=     3x = \frac{135}{20} - \frac{24}{20}

=     3x = \frac{111}{20}

=       x = \frac{111}{20} ÷ \frac{3}{1} = \frac{111}{20} × \frac{1}{3} = \frac{37}{20}

So, the weight of the equal bags are \frac{37}{20} and the weight of the third bag ( heavy one ) is \frac{37}{20} + \frac{6}{5} = \frac{37}{20} + \frac{24}{20} = \frac{61}{20}

1st bag =     \frac{37}{20} kg

2nd bag =  \frac{37}{20} kg

3rd bag =   \frac{61}{20} kg

Hope it helps...

7 0
3 years ago
(sin^4 theta - 1)/(cos^2 theta) = cos^2 theta - 2
WARRIOR [948]

Answer:

?

Step-by-step explanation:

3 0
2 years ago
PLEASE HELP. i’ll give brainliest if you want!
hammer [34]

Answer:

Dharma

Step-by-step explanation:

looked it up

3 0
3 years ago
Read 2 more answers
Problem<br> How many numbers between 111 and 100100100 (inclusive) are divisible by 101010 or 777?
olga2289 [7]

I suspect you meant

"How many numbers between 1 and 100 (inclusive) are divisible by 10 or 7?"

• Count the multiples of 10:

⌊100/10⌋ = ⌊10⌋ = 10

• Count the multiples of 7:

⌊100/7⌋ ≈ ⌊14.2857⌋ = 14

• Count the multiples of the LCM of 7 and 10. These numbers are coprime, so LCM(7, 10) = 7•10 = 70, and

⌊100/70⌋ ≈ ⌊1.42857⌋ = 1

(where ⌊<em>x</em>⌋ denotes the "floor" of <em>x</em>, meaning the largest integer that is smaller than <em>x</em>)

Then using the inclusion/exclusion principle, there are

10 + 14 - 1 = 23

numbers in the range 1-100 that are divisible by 10 or 7. In other words, add up the multiples of both 10 and 7, then subtract the common multiples, which are multiples of the LCM.

6 0
3 years ago
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