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Elenna [48]
2 years ago
11

Which fraction is equivalent to StartFraction 2 Over 6 EndFraction? StartFraction 3 Over 7 EndFraction, because StartFraction 2

Over 6 EndFraction = StartFraction 2 + 1 Over 6 + 1 EndFraction = StartFraction 3 Over 7 EndFraction StartFraction 3 Over 9 EndFraction, because StartFraction 2 Over 6 EndFraction = one-third and One-third = StartFraction 3 Over 9 EndFraction StartFraction 3 Over 12 EndFraction, because StartFraction 2 Over 6 EndFraction = one-third and One-third = StartFraction 3 Over 12 EndFraction StartFraction 3 Over 8 EndFraction, because StartFraction 2 Over 6 EndFraction = one-half = StartFraction 2 + 1 Over 6 + 2 EndFraction = StartFraction 3 Over 8 EndFraction
Mathematics
2 answers:
natka813 [3]2 years ago
8 0

Answer:

2/6

Step-by-step explanation:

irinina [24]2 years ago
7 0

Question:

Which fraction is equivalent to \frac{2}{6}?

- \frac{3}{7} because \frac{2}{6}= \frac{2 + 1}{6 + 1}

- \frac{3}{9} because \frac{2}{6}= \frac{1}{3} and \frac{1}{3} = \frac{3}{9}

- \frac{3}{12} because \frac{2}{6}= \frac{1}{3} and \frac{1}{3} = \frac{3}{9}

- \frac{3}{8} because \frac{2}{6}= \frac{1}{2} = \frac{2+1}{6+2} = \frac{3}{8}

Answer:

- \frac{3}{9} because \frac{2}{6}= \frac{1}{3} and \frac{1}{3} = \frac{3}{9}

Step-by-step explanation:

Two fractions are said to be equal if and only if they give the same value when simplified.

The equivalent of \frac{2}{6} is as explained in the selected option;

First, divide numerator and denominator by 2

\frac{2/2}{6/2}

Then simplify

2/2 = 1 and 6/2 = 3; So;

\frac{2/2}{6/2} = \frac{1}{3}

Multiply numerator and denominator by 3

\frac{1*3}{3*3} = \frac{3}{9}

Hence, \frac{2}{6} is equivalent to \frac{3}{9}

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A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?
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Answer:

(a) Selection = 6724520

(b) At\ most\ 12 = 6553976

(c) At\ most\ 8 = 6066720

(d) At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

Step-by-step explanation:

Given

Colors = 8

Balloons = 28 --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

n => 28 + 8-1= 35

r = 28

Selection = ^{35}^C_{28

Selection = \frac{35!}{(35 - 28)!28!}

Selection = \frac{35!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29}{7!}

Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}

Selection = \frac{33891580800}{5040}

Selection = 6724520

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

n => 15 + 8 -1 = 22

r = 15

Selection of at least 13 =

At\ least\ 13 = ^{22}C_{15}

At\ least\ 13 = \frac{22!}{(22-15)!15!}

At\ least\ 13 = \frac{22!}{7!15!}

At\ least\ 13 = 170544

Ways of selecting at most 12  =

At\ most\ 12 = Total - At\ least\ 13 --- Complement rule

At\ most\ 12 = 6724520- 170544

At\ most\ 12 = 6553976

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

n => 19+ 8 -1 = 26

r = 19

Selection of at least 9 =

At\ least\ 9 = ^{26}C_{19}

At\ least\ 9 = \frac{26!}{(26-19)!19!}

At\ least\ 9 = \frac{26!}{7!19!}

At\ least\ 9 = 657800

Ways of selecting at most 8  =

At\ most\ 8 = Total - At\ least\ 9 --- Complement rule

At\ most\ 8 = 6724520- 657800

At\ most\ 8 = 6066720

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

n =6+6-1 = 11

r = 6

Selection =

^{11}C_6 = \frac{11!}{(11-6)!6!}

^{11}C_6 = \frac{11!}{5!6!}

^{11}C_6 = 462

Using inclusion/exclusion rule of two sets:

Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)

Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462

Only\ 12\ red\ and\ only\ 8\ blue\ = 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  6724520 - 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

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370 College freshmen were interviewed. 200 were registered for an Advanced Math class. 270 were registered for an English class
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Answer:

Step-by-step explanation:

To aid in understanding, draw three circles...

100 were registered for both Math and English.

50 were registered for both Sport and English but not Math.

70 were registered for Math, Sport, and English.

From here we will be able to calculate for those that signed up for only English class: Total number of students that signed up for english class

a. none signed up only for Maths (refer to attached document)

b. How many signed up for neither Math nor English meaning they signed up for only sports which is 70 from the venn diagram in the attachment.

c. How many signed up for an English class and at least one other class?

From the venn diagram:

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we could also have an a English class and sports - 50

Thus total is 150.

d. What is the total fee for all the students enrolled in classes

From the diagram, 120 (english only and sport only) enrolled for only one class = 120 x $100 = $12,000

From the diagram, 180 (english and sport, maths and sport and english and maths) enrolled for two classes = 180 x $150 = $27,000

From the diagram, only 70 enrolled for the three classes = 70 x $200 = $14,000

Total fees = $12,000 + $27,000 + $14,000 = $53,000

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Answer:

It will take 4 hours for the second car to catch up with the first car.

Step-by-step explanation:

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A)

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Car 2 = 52/60 = 0.867 miles per minute

To check your calculation, Car 2 should be exactly twice as far

B) The street, as far as we know, is at least 104 miles

C) See calculation at the top - the first car is 52miles from the start line when the second car starts

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