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likoan [24]
3 years ago
7

(A LOT OF POINTS) Given the linear equation 2x + y = 6, perform the necessary operations to put the equation into the proper gen

eral form. Explain in complete sentences how you knew that the equation was in the proper general form. Complete your work in the space provided or upload a file that can display math symbols if your work requires it. Include the entire process for establishing the general form of the equation and the general form.

Mathematics
2 answers:
nlexa [21]3 years ago
8 0

Answer:

\huge\boxed{2x + y - 6 = 0}

Step-by-step explanation:

2x + y = 6

Subtracting both sides by 6

2x + y - 6 = 0

Comparing it with the general form of equation \sf Ax+By +C = 0 , we get:

A = 2, B = 1 and C = -6.

So, <u><em>the equation is in proper general form.</em></u>

weqwewe [10]3 years ago
8 0

Answer:

\boxed{2x+y-6=0}

Step-by-step explanation:

\sf The  \  general  \  form  \  for  \  the  \  equation  \  of  \  a  \  line  \  is  \  given \ as \ Ax+By+C=0.

2x+y=6

\sf Subtract \ 6 \ from \ both \ sides.

2x+y-6=6-6

2x+y-6=0

\sf A=2 \ \ \ B = 1 \ \ \ C=-6

\sf The \ equation \ is \ in \ general \ form.

\sf Graph \ equation:

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Pepsi [2]

Answer:

a) 95% confidence interval estimate of the true weight is (3.026, 3.274)

b) 99% confidence interval estimate of the true weight is (2.944, 3.356)

Step-by-step explanation:

Confidence Interval can be calculated using M±ME where

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  • ME is the margin of error from the mean

And margin of error (ME) can be calculated using the formula

ME=\frac{t*s}{\sqrt{N} } where

  • t is the corresponding statistic in the given confidence level and degrees of freedom(t-score)
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Using the numbers 95% confidence interval estimate of the true weight is:

3.150±\frac{2.776*0.1}{\sqrt{5} }≈3.150±0.124

And 99% confidence interval estimate of the true weight is:

3.150±\frac{4.604*0.1}{\sqrt{5} }≈3.150±0.206

3 0
3 years ago
Evalute the experssion -280+12x=
Alina [70]
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The average value of the function y = x² – 1 on [0, 12] is
shusha [124]

Answer:

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2) according to the rule above:

f_{avr}=\frac{1}{12-0} \int\limits^{12}_0 {(x^2-1)} \, dx=\frac{1}{12}(\frac{x^3}{3}-x)|^{12}_0=48-1=47.

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a

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(URGENT) need help with this so badly someone help asap
Fynjy0 [20]
The answer for A is 2,026. I set a porportion up which is 77/x = 3.8/100 

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