All categories

3 years ago

HELP ME PLEASE!!!!!!!!!!!GIVING BRAINLIEST!!!!!!

Mathematics

2 answers:

poizon [28]3 years ago

5 0

B. y = $\frac{1}{2}$ x + 1

Aleonysh [2.5K]3 years ago

3 0

Answer:

y = (1/2)x + 1 <---- Answer B

Step-by-step explanation:

In form y = mx + b, m is the slope and b is the y-intercept.

The y-intercept always has an x-coordinate of zero, swe are really asking what is the y-coordinate where the line intercepts the y-axis. Look at the picture. The coordinates where the line intercepts the y-axis are (x, y) = (0, 1).

So the y-intercept is 1.

b = 1

The slope is the rise (or fall) of the graph as it goes left to right.

We see it goes up 2 units (from y=1 to y=3) in the same interval where it goes to the right 4 units (from x=0 to x=4). So, the rise/run = +2/4 = 1/2.

m = 1/2

Just substitute the values we found for b and m back into "y=mx+b"

y = (1/2)x + 1 <---- Answer B

You might be interested in

A chemist examines 15 geological samples for potassium chloride concentration. The mean potassium chloride concentration for the

NikAS [45]

Answer:

$0.376-2.145\frac{0.0012}{\sqrt{15}}=0.375$

$0.376+2.145\frac{0.0012}{\sqrt{15}}=0.377$

And we are 95% confident that the true mean of chloride concentration is between 0.375 and 0.377 cc/ cubic meter

Step-by-step explanation:

Data provided

$\bar X=0.376$ represent the sample mean for the chloride concentration

$\mu$ population mean (variable of interest)

s=0.0012 represent the sample standard deviation

n=15 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since we need to find the critical value first we need to begin finding the degreed of freedom

$df=n-1=15-1=14$

The Confidence is 0.95 or 95%, the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , and the critical value for this case with a t distribution with 14 degrees of freedom is $t_{\alpha/2}=2.145$

And the confidence interval is:

$0.376-2.145\frac{0.0012}{\sqrt{15}}=0.375$

$0.376+2.145\frac{0.0012}{\sqrt{15}}=0.377$

And we are 95% confident that the true mean of chloride concentration is between 0.375 and 0.377 cc/ cubic meter

7 0

3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$