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joja [24]
2 years ago
6

Peter mixes 4 1/2 cups of orange juice, 2 1/4 of ginger ale and 6 1/3 cups of strawberry lemonade to make some punch. What is th

e total number of the cups of punch that peter makes
Mathematics
1 answer:
ruslelena [56]2 years ago
8 0
To add these amounts together, we must first find their least common multiple in order to get common denominators (b/c when you add fractions, the denominators must be the same).

We'll start by listing some of their multiples.
To do this, count by whatever the denominator is:
4 1/2 (denominator is 2): 2 4 6 8 10 12 14
2 1/4 (denominator is 4): 4 8 12 16
6 1/3 (denominator is 3): 3 6 9 12 15

Look and see which is the first multiple that all three denominators have. Circle them if it helps you. In this case, it's 12.

So now we have to multiply the denominators by whatever number it takes to reach 12, and multiply by the same number to the numerator:

4 1/2 (times 6 to both top and bottom) =
4 6/12

2 1/4 (times 3) = 2 3/12

6 1/3 (times 4) = 6 4/12

Add all these fractions together, and you get 12 13/12, which is equal to 13 1/12.

Thus, Peter makes a total of 13 1/2 cups.

Hope this made sense! tell me if anything is confusing/incorrect :))
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Minchanka [31]
I have to think this again. be back later. X=2.22
3 0
2 years ago
In a class of 30 students (x+10) study algebra, (10x+3) study statistics, 4 study both algebra and statistics. 2x study only alg
Vladimir [108]

Answer:

1. The Venn diagrams are attached

2. When the statistics students number = 10·x + 3, we have;

The number of students that study

a. Algebra = 128/11

b. Statistic = 213/11

When the statistics students number = 2·x + 3, we have;

The number of students that study

a. Algebra = 16

b. Statistic = 15

Step-by-step explanation:

The parameters given are;

Total number of students = 30

Number of students that study algebra n(A) = x + 10

Number of students that study statistics n(B) = 10·x + 3

Number of student that study both algebra and statistics n(A∩B) = 4

Number of student that study only algebra n(A\B) = 2·x

Number of students that study neither algebra or statistics n(A∪B)' = 3

Therefore;

The number of students that study either algebra or statistics = n(A∪B)

From set theory we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 10·x + 3 - 4 = 27

11·x+13 = 27 + 4 = 31

11·x = 18

x = 18/11

The number of students that study

a. Algebra

n(A) = 18/11 + 10 = 128/11

b. Statistic

n(B) = 213/11

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11

However, assuming n(B) = (2·x + 3), we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 2·x + 3 - 4 = 27

2·x+3 + x + 10= 27 + 4 = 31

3·x = 18

x = 6

Therefore, the number of students that study

a. Algebra

n(A) = 16

b. Statistics

n(B) = 15

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11

The Venn diagrams can be presented as follows;

6 0
2 years ago
I round to 900 I have twice as many hundreds as ones and all my digits are different but their sums are 18 can someone answer th
stellarik [79]

Answer:

The number of once is 9.1  

The number of hundreds is 8.9

Step-by-step explanation:

Given as :

The total of digits having ones and hundreds = 900

The sum of digits = 18

Let The number of ones digit = O

And The number of hundreds digit = H

So, According to question

H + O = 18             .........1

100 × H + 1 × O = 900          ........2

Solving the equation

( 100 × H - H ) + ( O - O ) = 900 - 18

Or, 99 H + 0 = 882

Or , 99 H = 882

∴   H = \frac{882}{99}

I.e H = 8.9

Put the value of H in eq 1

So, O = 18 - H

I.e  O = 18 - 8.9

∴    O = 9.1

So, number of once = 9.1

number of hundreds = 8.9

Hence The number of once is 9.1  and The number of hundreds is 8.9

Answer

4 0
2 years ago
Pablo and Nadia are playing a game on a coordinate grid. Pablo puts a playing piece at point (-2, 3) and Nadia puts a playing pi
7nadin3 [17]

Answer:

The distance between the two playing pieces is of 6.4 units.

Step-by-step explanation:

Suppose we have two points:

A = (x_{1}, y_{1})

B = (x_{2}, y_{2})

The distance between these points is:

D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Distance between:

A(-2,3) and B(3,-1)

D = \sqrt{(3 - (-2))^{2} + (-1 - 3)^{2}} = \sqrt{41} = 6.4

The distance between the two playing pieces is of 6.4 units.

5 0
2 years ago
What is the greatest common factor of 16x4y3 and 12x2y7?
Marta_Voda [28]

The greatest common factor, or GCF, is the greatest factor that divides two numbers. To find the GCF of two expression

<span>1.     </span>List the prime factors of each number.

<span>2.     </span>Multiply those factors both numbers have in common. If there are no common prime factors, the GCF is 1.

16x^4y^3 = ( 4*4)(x*x*x*x)(y*y*y)

12x^2y^7 = (4*3)(x*x)(y*y*y*y*y*y*y)

<span>So common to them is 4x^2y^3 </span>

4 0
2 years ago
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