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4vir4ik [10]
3 years ago
8

you have a $100 gift card to spend at a store. You buy a portable compact disc player for $45. Compact discs are on sale for $11

each. How may compact discs can you buy with the money remaining on gift card?
Mathematics
2 answers:
kotegsom [21]3 years ago
8 0
Let's create an expression with the given variables and values to better understand the problem.

y=100-(45+11c)

This equation subtracts the value of the cd player and the CDs from the money on the gift card. 

We can simplify the equation a little more to make it easier to solve.
y=55-11c

Now, we can see how many CDs we can buy with the remaining money on the gift card by solving for 'c'.
11c=55
c=5

We can buy 5 CDs with the money remaining on the gift card.
abruzzese [7]3 years ago
7 0
You can buy 4 compact discs
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Class of 50 students dressed up 38% have white shirts. How many are wearing white shirts?
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Step-by-step explanation:

☥ \underline{ \underline{ \text{Given} }}:

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☥ \underline{ \underline{ \sf{Solution} }} :

⟹ \sf{38\% \: of \: 50} =  \frac{38}{100}  \times 50 =  \boxed{ \sf{19}}

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3 0
3 years ago
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

=k^1+2k+1+k+1

=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

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Answer:

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So:  any score that lies between 88.8 and 97.2 is within one std. dev. of the mean.

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