5y+6+7y=?(6y+?)what is the answer

The graphs below are both absolute value functions. The equation of the redgraph is f(x) = |xl. Which of these is the equation o

grigory [225]

ANSWER

$\Rightarrow y=\frac{1}{2}|x|$

EXPLANATION

We want to find the absolute value function for the line in blue.

The general form of an absolute value function is:

$y=a|x-h|+k$

where (h, k) = vertex

From the line, the vertex of the graph in blue is:

$(0,0)$

To find a, we have to pick a point (x, y) on the line and input it into the general function.

Let us pick (2, 1).

Therefore, we have:

$\begin{gathered} 1=a|2-0|+0 \\ 1=a|2|=2a \\ \Rightarrow a=\frac{1}{2} \end{gathered}$

Therefore, the absolute value function is:

$\begin{gathered} y=\frac{1}{2}|x-0|+0 \\ \Rightarrow y=\frac{1}{2}|x| \end{gathered}$

3 0

1 year ago

What is the equation in point-slope form of a line that passes through the points (−4, −1) and (5, 7) ?

Sergio039 [100]

The point-slope form:

$y-y_1=m(x-x_1)$

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-4, -1) and (5, 7). Substitute:

$m=\dfrac{7-(-1)}{5-(-4)}=\dfrac{7+1}{5+4}=\dfrac{8}{9}\\\\\boxed{y-7=\dfrac{8}{9}(x-5)}$

6 0

3 years ago

GIVING THE BRAINLIEST ANSWER

Ahat [919]

Answer:

(6,3)

Step-by-step explanation:

y=2/3 x - 1

y=-1/2 x + 6

Since both equations are equal to y, we can set them equal

2/3 x - 1 =-1/2 x + 6

We have fractions, so I will multiply by 6 to clear the fractions

6(2/3 x - 1) =(-1/2 x + 6)6

Distribute

4x -6 = -3x +36

Add 3x to each side

4x+3x -6 = -3x+3x +36

7x -6 = 36

Add 6 to each side

7x-6+6 = 36+6

7x = 42

Divide each side by 7

7x/7 = 42/7

x =6

Now we need to find y

y =2/3x -1

y = 2/3(6) -1

y = 4-1

y=3

(6,3)

5 0

3 years ago

Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The

Marina86 [1]

Answer:

Step-by-step explanation:

Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

G = card drawn is green

Y = card drawn is yellow

E = card drawn is even-numbered

List:

Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}

2) P(G) = 5/8

3) P(G/E) = P(GE)/P(E)

GE = {G2, G4}

Hence P(G/E) = 2/5

4) GE = {G2, G4}

P(GE) = 2/8 = 1/4

5) P(G or E) = P(G)+P(E)-P(GE)

= 5/8 + 3/8-2/8 = 3/5

6) No there is common element as G2 and G4

Cannot be mutually exclusive

6 0

3 years ago

. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting

SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, $\overline x_1$ = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, $\overline x_2$ = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20 - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; $t_{\alpha /2}$ = ±2.026192

The t-confidence interval is given as follows;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

Therefore;

$\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}$

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0

3 years ago