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Maurinko [17]
4 years ago
10

Jaime has 5 over 11 spaceof a project completed while Tim has finished 7 over 13 spaceof the same project.

Mathematics
1 answer:
viva [34]4 years ago
8 0

Answer:

Part a) Tim has completed the greater amount of work

Part b) They have completed together \frac{142}{143} of the project or 99.3\% of the project

Step-by-step explanation:

Part a) Who has completed the greater amount of work?

we have

Jaime has 5/11 of a project completed

Tim has 7/13 of a project completed  

Multiply 5/11 by 13/13 and 7/13 by 11/11

so

Jaime ---> \frac{5}{11}(\frac{13}{13})=\frac{65}{143}

Tim ---> \frac{7}{13}(\frac{11}{11})=\frac{77}{143}

Remember that

When fractions have the same denominator, the larger fraction is the one with the larger numerator  

\frac{77}{143} >\frac{65}{143}

therefore

Tim has completed the greater amount of work

Part b)  How much of the project have they completed together?

we have that

Jaime has 65/143 of a project completed  

Tim has 77/143 of a project completed  

Sum the fraction of the project completed by Jaime plus the fraction of the project completed by Tim

\frac{65}{143}+\frac{77}{143}=\frac{142}{143}

Convert to percentage

Multiply by 100

\frac{142}{143}(100)=99.3\%

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Joe has one book each for algebra, geometry, history, psychology, Spanish, English and Physics in his locker. How many different
Alisiya [41]

Answer:

There are 35 different sets of 3 books Joe could choose

Step-by-step explanation:

* Lets explain how to solve the problem

- Combination is a collection of the objects where the order doesn't

 matter

- The formula for the number of possible combinations of r objects from

 a set of n objects is nCr = n!/r!(n-r)!

- n! = n(n - 1)(n - 2)................. × 1

Lets solve the problem

- Joe has one book each for algebra, geometry, history, psychology,

 Spanish, English and Physics in his locker

∴ He has <em>seven</em> books in the locker

- He wants to chose <em>three</em> of them

∵ The order is not important when he chose the books

∴ We will use the combination <em>nCr</em> to find how many different sets

  of three books he can choose

- The total number of books is 7

∴ n = 7

∵ He chooses 3 of them

∴ r = 3

∵ 7C3 = 7!/3!(7 - 3)! = 7!/3!(4!)

∴ 7C3=\frac{(7)(6)(5)(4)(3)(2)(1)}{[(3)(2)(1)][(4)(3)(2)(1)]}=35

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3 0
3 years ago
Anybody knows what the answer is
Lady bird [3.3K]
Im pretty sure it would be 4^-2 because when you’re multiplying two different exponents you add them. i hope i helped!
7 0
3 years ago
12x -20=n (3x-5)<br><br> I need help I don’t get it please help me.
Salsk061 [2.6K]

Answer:

The value of n = 4 and  

The value of x = \frac{5}{3}

Step-by-step explanation:

Method I:

As given, 12x -20=n (3x-5)      .........(1)

Firstly , take 12x - 20

As 12x - 20 can be written as 4(3x - 5)

∴ equation (1) becomes

4(3x - 5) = n (3x-5)

Divide the equation by (3x-5) , we get

4 = n

Method II:

As given, 12x -20=n (3x-5)

⇒12x - 20 = 3nx - 5n

⇒12x - 3nx = 20 - 5n

⇒3x(4 - n) = 5( 4 - n)

Divide the equation by (4-n) , we get

3x = 5

⇒ x = \frac{5}{3}

∴ we get

n = 4 and  x = \frac{5}{3}

Now, check that value of n and x satisfies the equation or not

12(\frac{5}{3}) - 20 = 4( 3(\frac{5}{3}) - 5)

⇒4(5) - 20 = 4(5 - 5)

⇒20 - 20 = 4(0)

⇒ 0 = 0

Hence , we get

The values of x and n satisfies the equation

so, we get

n = 4 and  x = \frac{5}{3}

5 0
3 years ago
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