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Furkat [3]
3 years ago
14

A candy store sells 8-pound bags of mixed hazelnuts and cashews. If c pounds of cashews are in a bag, the price p of the bag can

be found using the formula p= 2.59c + 1.72(8-c). If one bag is priced at $18.11, how many pounds of cashews does it contain
Mathematics
1 answer:
Butoxors [25]3 years ago
4 0
C= how many pounds of cashews in a bag
P=how much the cost
Solve for C
we know that P would have to be $18.11
Sub wat P= to in the equation
$18.11=2.59c+1.79(8-c)
Start with multiplying
1.79(8)=14.32(c)=14.32c
Then add like terms
2.59c+14.32c=16.91c
Divide to get C by itself. So that 16.91 cancel itself.
16.91c/16.91c=18.11/16.91
C=1.07
I might be wrong but I tried so wateves

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A) How many ways can 2 integers from 1,2,...,100 be selected
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→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

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(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

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