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Juliette [100K]
2 years ago
8

The equation of the linear regression line represents the relationship between the amount a person earns, x, and the amount save

d, y.
yˆ=0.01x+50

What does the slope of the line represent?



For every $100 a person earns, the savings increases by 1 cent.

For every $100 a person earns, the savings increases by $1.

For every $100 that a person earns, the savings decreases by $1.

For every $1 a person earns, the savings increases by $100.
Mathematics
2 answers:
koban [17]2 years ago
6 0

Answer:

Option B is the correct option.

Step-by-step explanation:

The equation of the linear regression line represents the relationship between the amount and a person's earning.

y = .01x + 50

Or y=\frac{1}{100}x+50

It's in the form of y = mx + c

Where y = amount of saving

x = amount of saving

m = slope of the line = change in saving/change in earning

m = 1/100 = saving increase by $1 / earning increase of $100

Therefore slope represents for every $100 a person earns, the saving increases by $1.

Option B. is the answer.  

givi [52]2 years ago
3 0
Hi! Did you get the answer for this by any chance??
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2 years ago
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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The same applies to benchmark numbers in intervals of 100. If you want to find 170, used the benchmark numbers 100 and 200. Then, you estimate at which point represents 170. For 410, you base on the benchmark numbers 400 and 500.

6 0
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Read 2 more answers
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