Explain how you determine an equation for a line using a table of input/output values.

1 answer:

6 0

Example :

x y
1 3
2 6
3 9
4 12

first thing u do is pick any 2 points (x,y) from ur table

(1,3) and (2,6)
now we sub those into the slope formula (y2 - y1) / (x2 - x1) to find the slope
(y2 - y1) / (x2 - x1)
(1,3)....x1 = 1 and y1 = 3
(2,6)...x2 = 2 and y2 = 6
sub
slope = (6 - 3) / (2 - 1) = 3/1 = 3

now we use slope intercept formula y = mx + b
y = mx + b
slope(m) = 3
use any point off ur table...(1,3)...x = 1 and y = 3
now we sub and find b, the y int
3 = 3(1) + b
3 = 3 + b
3 - 3 = b
0 = b

so ur equation is : y = 3x + 0....which can be written as y = 3x...and if u sub any of ur points into this equation, they should make the equation true....if they dont, then it is not correct

and if u need it in standard form..
y = 3x
-3x + y = 0
3x - y = 0 ...this is standard form

You might be interested in

How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$

where $|Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}$

and

$|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}$

What we are looking for is the number of permutations of the 26 letters of the alphabet that do not contain the strings fish, rat or bird, or

$|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}$

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0

3 years ago

g You are planning an opinion study and are considering taking an SRS of either 200 or 600 people. Explain how the sampling dist

Katen [24]

Answer:

By the Central Limit Theorem, both would be approximately normal and have the same mean. The difference is in the standard deviation, since as the sample size increases, the standard deviation decreases. So the SRS of 600 would have a smaller standard deviation than the SRS of 200.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .