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VashaNatasha [74]
2 years ago
15

HELP ME PLZ DONT USE DEMOS GRAPHING graph y=-1/4x+6​

Mathematics
2 answers:
slava [35]2 years ago
8 0

Answer:

The line would contain (0,6) and (-4,7).

Step-by-step explanation:

The equation is in slope intercept form.  

y = mx + b\\\rule{150}{0.5}\\m - \text{slope}\\b - \text{y - intercept.}

'6' takes b's spot, so we would plot a point at (0,6).

The line has a slope of negative one-fourth. From the intercept, we would move 4 units left and the move up one.

We would now be at (-4,7). With the two points, we would connect the two points together to make the line.

Hope this helps.

alexdok [17]2 years ago
5 0

Answer:

.

Step-by-step explanation:

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The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu
AURORKA [14]

Answer:

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21        x2`= 27

n1=  10       n2= 14

s1= 5.6       s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27)  +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 <  ( 6 )  +2.11*2.111

6- 4.4521 < u1-u2 <  ( 6 )  +5.294

- 1.5479 < u1-u2 <  10.4521

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`)  /√s1²/n1 + s2²/n2

t= (21-27)  /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

5 0
3 years ago
(x^(2)-11x+30)/(x^(2)-25)*(x^(2)+6x+5)/(x-5x-6) simplify the rational expression
DanielleElmas [232]
Simplify the following:
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The factors of 5 that sum to 6 are 5 and 1. So, x^2 + 6 x + 5 = (x + 5) (x + 1):
((x + 5) (x + 1) (x^2 - 11 x + 30))/((x^2 - 25) (x - 5 x - 6))

The factors of 30 that sum to -11 are -5 and -6. So, x^2 - 11 x + 30 = (x - 5) (x - 6):
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (x - 5 x - 6))

x - 5 x = -4 x:
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (-4 x - 6))

Factor -2 out of -4 x - 6:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (2 x + 3) (x^2 - 25))

x^2 - 25 = x^2 - 5^2:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (x^2 - 5^2) (2 x + 3))

Factor the difference of two squares. x^2 - 5^2 = (x - 5) (x + 5):
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((x - 5) (x - 6) (x + 5) (x + 1))/((x - 5) (x + 5) (-2) (2 x + 3)) = ((x - 5) (x + 5))/((x - 5) (x + 5))×((x - 6) (x + 1))/(-2 (2 x + 3)) = ((x - 6) (x + 1))/(-2 (2 x + 3)):
((x - 6) (x + 1))/(-2 (2 x + 3))

Multiply numerator and denominator of ((x - 6) (x + 1))/(-2 (2 x + 3)) by -1:

Answer: (-(x - 6) (x + 1))/(2 (2 x + 3))
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