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Zielflug [23.3K]
3 years ago
6

The florist makes the greatest number of identical arrangements with the carnations and asters. She has 72 carnations and 42 ast

ers. How can she decide how many carnations to place in each arrangement?
A)Subtract the number of asters from the number of carnations.
B)Divide the number of carnations by the GCF of 72 and 42.
C)Find the least common multiple of 72 and 42. D)Find the greatest common factor of 72 and 42.
Mathematics
1 answer:
Studentka2010 [4]3 years ago
7 0

the answer is B. Hope this helps

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The first digit of the quotient is the first number of the answer of your division problem. the answer is 2
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PLEASEEEEE HELP BIT FINALS TEST!!!! PLSSS :(
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The 1st, 3rd, and the 4th

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Kenji is decorating papers with a total of 15 heart stickers and 12 star stickers for the child he is babysitting. If he wants t
Marina CMI [18]

Answer: The greatest number of pages Kenji can decorate = 3

Step-by-step explanation:

Given: Total heart stickers = 15

Total star stickers =12

If all the papers identical, with the same combination of heart and star stickers and no stickers left over.

Then the greatest number of pages Kenji can decorate = GCD(15,12)  [GCD=greatest common divisor]

Since 15 = 3 x 5

12=2 x 2 x 3

GCD(15,12) =3

Hence, the greatest number of pages Kenji can decorate = 3

7 0
3 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
What's is greater 34.653 or 4.653?
Strike441 [17]
The greater one is 34.653
because

34 is greater than 4
 so, 34.653>4.653

4 0
3 years ago
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