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shusha [124]
4 years ago
10

If Samantha wants two bags of chips and a coke,how much should she plan to spend?

Mathematics
2 answers:
LuckyWell [14K]4 years ago
7 0

Answer:

C. 7$

Step-by-step explanation:

kotykmax [81]4 years ago
5 0

Answer:

Samantha should plan to spend $7.

Step-by-step explanation:

Let the price of 1 pizza be = p

Let the price of 1 coke = c

Let the price of 1 pack of chips = b

As per table, we get following equations:

p+c+b=9 or p=9-c-b   ......(1)

p+2c=10     .......(2)

2p+2b=12    ......(3)

Substituting the value of p from (1) in (2)

9-c-b+2c=10

=> c-b=1    ......(4)

Substituting the value of p from (1) in (3)

2(9-c-b)+2b=12

=> 18-2c-2b+2b=12

=> 18-2c=12

=> 2c=18-12

=> 2c=6

c = 3

We have c-b=1

=> b=c-1

=> b=3-1

b = 2

And p=9-c-b

p=9-3-2

p = 4

We get the following cost now.

Cost of 1 pizza = $4

Cost of 1 coke = $3

Cost of 1 chips bag = $2

We have been given that Samantha wants two bags of chips and a coke.

So, she should spend (2\times2)+3= 4+3=7

Hence, Samantha should plan to spend $7.

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Of the students in the college, 60% of the students reside in the hostel and 40% of the students are day scholars. Previous year
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Answer:

the probability that the student is a hostlier is 0.6923

Step-by-step explanation:

The computation of the probability that the student is a hostlier is shown below

= (0.60 × 0.30) ÷ (0.60 × 0.30) + (0.40 × 0.20)

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Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6
musickatia [10]

Answer:

(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

Step-by-step explanation:

In order to find the values of (x+y)^5 and (x+y)^6, you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

(x+y)^5 = \sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i = \frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5 = x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5.

(x+y)^6 = \sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i = \frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6= x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6.

5 0
3 years ago
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