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Brrunno [24]
3 years ago
5

Find the greatest common factor of 26 and 14​

Mathematics
2 answers:
Jlenok [28]3 years ago
8 0
The gcf is 2 hope this helps
Nookie1986 [14]3 years ago
6 0
The greatest common factor of 26 and 14 is 2.
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12/25 * 100 = 48%

Amount/Whole Amount * 100

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Omg please help ASAP!!
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try the first, fourth, and last one

Step-by-step explanation: when you read the question it say select all that apply to form the equation so all you have to do is select the numbers and the variables that are in the equation above.

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3 years ago
Juarez scored 32 points in a basketball game. He scored twice as many baskets worth 3 points than baskets worth 2 points. How ma
Vladimir79 [104]

Answer:

8 baskets worth 3 points each and 4 baskets worth 2 points each

Step-by-step explanation:

It is 8 baskets worth 3 points each and 4 baskets worth 2 points each because it says Juarez scored twice as many baskets worth 3 points than the baskets worth 2 points. 4 x 2 = 8

-----------------------------------------------------------------------------

6 baskets worth 3 points each and 7 baskets worth 2 points each : wrong because 7 x 2 = 14.

10 baskets worth 3 points each and 1 basket worth 2 points : wrong because 1 x 2 = 2.

4 baskets worth 3 points each and 10 baskets worth 2 points each: wrong because 10 x 2 = 40.

8 0
3 years ago
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Determine whether the equation below has a one solutions, no solutions, or an
lesya [120]

Answer:

I don't have the idea of this answers

8 0
3 years ago
PLEASE HELP ME
denis23 [38]

Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

      (2, 1), (2, 2), (2, 3), (2, 4)

      (3, 1), (3, 2), (3, 3), (3, 4)

      (4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24

Thus, the number of times should Hartley spin a difference of 0 is 24.

4 0
3 years ago
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