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Marianna [84]
4 years ago
5

a store sells small notebooks for $7 and large notebooks for $10. If a student buys 6 notebooks and spends $51, how many of each

size did he buy?
Mathematics
2 answers:
____ [38]4 years ago
4 0
3 small ones and 3 big ones because 3*7=21 plus 10*3=20 =51
andreyandreev [35.5K]4 years ago
3 0

Answer:

3 small notebooks and 3 large notebooks

Step-by-step explanation:

The cost spent by the student is the sum of the cost spent on small notebooks and the cost spent on large notebooks.

The cost spent on small note books is the product of the cost per notebook and the number of small note books purchased.

let the number of small and large notebooks purchased be a and b respectively,

a + b = 6

and

7a + 10b = 51

since a + b = 6,

b = 6 - a

7a + 10 (6 - a) = 51

7a + 60 - 10a = 51

-3a = 51 - 60

-3a = -9

a = 3

since b = 6 - a,

b = 6 - 3

= 3

Hence he bought 3 small notebooks and 3 large notebooks.

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3 years ago
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What is the area of the parallelogram if each square is 1 square foot? a. 12 ft ³ b. 12 ft ² c. 12 d. 12 feet
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Answer:

12 ft² :)

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3 years ago
Assume that 12 people, including the husband and wife pair, apply for 4 sales positions. People are hired at random.
Iteru [2.4K]

brainly.com/question/5218999


The formula C(n, r)= \frac{n!}{r!(n-r)!}, where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

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


Selecting 4 people out of 12 can be done in :

\displaystyle{C(12, 4)= \frac{12!}{4!8!}= \frac{12\cdot11\cdot10\cdot9\cdot8!}{4!8!}= \frac{12\cdot11\cdot10\cdot9}{4!}=11\cdot5\cdot9= 495       many ways.


All the possible groups of 4 people, where the husband and wife are included, can be done in C(10, 2) many ways, since we only calculate the possible choices of 2 out of 10 people, to complete the groups of 4.


\displaystyle{ C(10, 2)= \frac{10!}{2!8!}= \frac{10\cdot9}{2}=45


Thus, the 

probability that both the husband and wife are hired is 45/495=0.09


Part 2)

The probability that one is hired and the other is not = 

P(husband hired, wife not hired) + P(wife hired, husband not hired)

these 2 are clearly equal, so it is enough to calculate one.


Consider the case : husband hired, wife not hired.

assuming the husband is hired, we have to calculate the possible groups of 3 that can be formed from 11-1 (the wife)=10 people.

this is 

\displaystyle{ C(10, 3)= \frac{10!}{3!7!}= \frac{10 \cdot9 \cdot8}{3\cdot2}=10\cdot3\cdot4=120


thus, 


P(husband hired, wife not hired)=120/495=0.24


thus, 

The probability that one is hired and the other is not = 

P(husband hired, wife not hired) + P(wife hired, husband not hired) =

0.24+0.24=0.48



Answer:


A) 0.09


B) 0.48

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