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Nataliya [291]
2 years ago
15

Another florist has 16 yellow mums and 40 white carnations. Which question can be answered by finding the GCF of 16 and 40? How

many different combinations of yellow mums and white carnations can the florist make? What is the greatest number of identical groups that can be made with a number of mums and carnations in each group? How many total number of flowers will be in each group of mums and carnations? What is the greatest number of white carnations that can be in each group?
Mathematics
2 answers:
Zolol [24]2 years ago
8 0

Answer:What is the greatest number of identical groups that can be made with a number of mums and carnations in each group?

Step-by-step explanation:

EZ

levacccp [35]2 years ago
6 0

Answer:

B

Step-by-step explanation: I got correct on ed2020!

Hope this helps!~ :>

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Compute the mode, median, and mean for 23455
Katarina [22]

See picture for math work and answer.

7 0
3 years ago
Solve x - 3 1/8 = 5/12
Anna007 [38]

Answer:

\large\boxed{x=\dfrac{85}{24}=3\dfrac{13}{24}}

Step-by-step explanation:

Convert the mixed number to the fraction:

3\dfrac{1}{8}=\dfrac{3\cdot8+1}{8}=\dfrac{25}{8}

METHOD 1:

x-3\dfrac{1}{8}=\dfrac{5}{12}\\\\x-\dfrac{25}{8}=\dfrac{5}{12}\\\\\text{the common denominator is 24}\\\\x-\dfrac{25}{8}=\dfrac{5}{12}\qquad\text{multiply both sides by 24}\\\\24x-(3)(25)=(2)(5)\\\\24x-75=10\qquad\text{add 75 to both sides}\\\\24x=85\qquad\text{divide both sides by 24}\\\\\boxed{x=\dfrac{85}{24}}

METHOD 2:

x-3\dfrac{1}{8}=\dfrac{5}{12}\qquad\text{add}\ 3\dfrac{1}{8}\ \text{to oboth sides}\\\\x=\dfrac{5}{12}+3\dfrac{1}{8}\\\\x=\dfrac{5\cdot2}{12\cdot2}+3\dfrac{1\cdot3}{8\cdot3}\\\\x=\dfrac{10}{24}+3\dfrac{3}{24}\\\\x=3\dfrac{10+3}{24}\\\\\boxed{x=3\dfrac{13}{24}}

7 0
3 years ago
A rectangle has length x cm and width 2 cm less
vichka [17]

x^2 - 2x = 146

x^2 - 2x - 146 = 0

See attachment. The quadratic formula is needed in this case.

7 0
3 years ago
Read 2 more answers
Points Points Points
alex41 [277]
Yesyesyes thank you have an amazing day!
3 0
3 years ago
Read 2 more answers
1. Consider the data set 1,2,3,4,5,6,7,8,9.
ivanzaharov [21]

Answer:  a) Mean = 5, Median = 5

b) Mean = 15, Median = 5

c) Due to presence of outlier i.e. 99.

Step-by-step explanation:

Since we have given that

1,2,3,4,5,6,7,8,9

Here, n = 9 which is odd

So, Mean would be

\dfrac{1+2+3+4+5+6+7+8+9}{9}=\dfrac{45}{9}=5

Median = (\dfrac{n+1}{2})^{th}=\dfrac{9+1}{2}=5^{th}=5

If 9 is replaced by 99,

1,2,3,4,5,6,7,8,99

So, mean would be

\dfrac{1+2+3+4+5+6+7+8+99}{9}=\dfrac{135}{9}=15

Median would be same as before i.e. 5

The mean is neither central nor typical for the data due to outlier i.e. 99

Hence, a) Mean = 5, Median = 5

b) Mean = 15, Median = 5

c) Due to presence of outlier i.e. 99.

5 0
3 years ago
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