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2 years ago

Cheese sticks that were previously priced at "10 for $1" are now "2 for $1". Find each percent change.

Mathematics

1 answer:

AysviL [449]2 years ago

7 0

Answer:

The percent decrease in the number of cheese sticks you can buy for $1 is 80%.

Step-by-step explanation:

Given : Cheese sticks that were previously priced at "10 for $1" are now "2 for $1".

To find : The percent decrease in the number of cheese sticks you can buy for $1 ?

Solution :

The formula used to find percent decrease is given by,

$\%\text{ change}=\frac{\text{Amount of change}}{\text{Original amount }}\times 100$

The price change from 10 to 2,

$\%\text{ change}=\frac{10-2}{10}\times 100$

$\%\text{ change}=\frac{8}{10}\times 100$

$\%\text{ change}=80\%$

The percent decrease in the number of cheese sticks you can buy for $1 is 80%.

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Xelga [282]

Answer:

D. 63%

Step-by-step explanation:

the white marble was picked out 41 of the 65 times, so 41/65

divide 41 by 65 in your calculator and you get .6307

then multiply it by 100 because it asked for a percentage and you get 63%

Ta-Da!

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Answer:

Answer D is correct

Step-by-step explanation:

1: (1, 1.25) (2, 2)

So you have your first problem, your first number is x1 while your next number in the first set of parenthesis is y1. Your next set of parenthesis will be x2 and y2 like this:

x1 y1 x2 y2

(1, 1.25) (2, 2)

Then you set up a equation like this!

x2-x1

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y2-y1

so we now plug in the numbers and get this

2-1.25 = 0.75

--------- ---- or 0.75 BUT not 1.25 like we need!

2-1 = 1

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2 years ago

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Last year Ron bought 40 pen sets and sold 25. This year he bought 75 pen sets and sold 60. The absolute error each year was the

vagabundo [1.1K]

Answer:200

Step-by-step explanation:

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1 year ago

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

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