What is the equation of the line of best fit?

Use properties to rewrite the given equation. Which equations have the same solution as 3/5x +2/3 + x = 1/2– 1/5x? Check all tha

vodomira [7]

we have

$\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x$

Combine like terms in both sides

$(\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

we know that

$(\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x$

substitute in the expression above

$\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$ -----> equation A

Multiply equation A by $5*3*2=30$ both sides

$30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)$

$48x+20=15-6x$ ---------> equation B

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$48x+6x=15-20$

$54x=-5$ ---------> equation C

Solve for x

$x=-\frac{5}{54} =-0.09$

We are going to proceed to verify each case to determine the solution.

Case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

the case a) is equal to the equation A

so

the case a) have the same solution that the given equation

Case b) $18x+20+30x=15-6x$

Combine like terms in left side

$(18x+30x)+20=15-6x$

$(48x)+20=15-6x$

the case b) is equal to the equation B

so

the case b) have the same solution that the given equation

Case c) $18x+20+x=15-6x$

Combine like terms in left side

$(18x+x)+20=15-6x$

$(19x)+20=15-6x$

$19x+6x=15-20\\25x=-5\\x=-0.20$

$-0.20\neq -0.09$

therefore

the case c) not have the same solution that the given equation

Case d) $24x+30x=-5$

Combine like terms in left side

$54x=-5$

the case d) is equal to the equation C

so

the case d) have the same solution that the given equation

Case e) $12x+30x=-5$

Combine like terms in left side

$42x=-5$

$x=-5/42=-0.12$

$-0.12\neq -0.09$

therefore

the case e) not have the same solution that the given equation

therefore

the answer is

case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

case b) $18x+20+30x=15-6x$

case d) $24x+30x=-5$

7 0

4 years ago

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50 points for the person to answer these :)

svetoff [14.1K]

I'm assuming you don't still need these? Sorry I didn't see this in time

7 0

3 years ago

The following table shows the number of innings pitched by each of the Greenbury Goblin's starting pitchers during the Rockbotto

Sedaia [141]

Answer:

Step-by-step explanation

i can help u, but what are the numbers

7 0

4 years ago

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An engineer on the ground is looking at the top of a building the angle of elevation to the top of the building is 22 degrees th

vekshin1

Hello,

tan 22°=450/d
==>d=450/tan 22°=1113.7890... ≈1114 (ft)

7 0

3 years ago

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In a certain community, eight percent of all adults over age 50 have diabetes. If a health service in this community correctly d

____ [38]

Complete question is;

In a certain community, 8% of all people above 50 years of age have diabetes. A health service in this community correctly diagnoses 95% of all person with diabetes as having the disease, and incorrectly diagnoses 10% of all person without diabetes as having the disease. Find the probability that a person randomly selected from among all people of age above 50 and diagnosed by the health service as having diabetes actually has the disease.

Answer:

P(has diabetes | positive) = 0.442

Step-by-step explanation:

Probability of having diabetes and being positive is;

P(positive & has diabetes) = P(has diabetes) × P(positive | has diabetes)

We are told 8% or 0.08 have diabetes and there's a correct diagnosis of 95% of all the persons with diabetes having the disease.

Thus;

P(positive & has diabetes) = 0.08 × 0.95 = 0.076

P(negative & has diabetes) = P(has diabetes) × (1 –P(positive | has diabetes)) = 0.08 × (1 - 0.95)

P(negative & has diabetes) = 0.004

P(positive & no diabetes) = P(no diabetes) × P(positive | no diabetes)

We are told that there is an incorrect diagnoses of 10% of all persons without diabetes as having the disease

Thus;

P(positive & no diabetes) = 0.92 × 0.1 = 0.092

P(negative &no diabetes) =P(no diabetes) × (1 –P(positive | no diabetes)) = 0.92 × (1 - 0.1)

P(negative &no diabetes) = 0.828

Probability that a person selected having diabetes actually has the disease is;

P(has diabetes | positive) =P(positive & has diabetes) / P(positive)

P(positive) = 0.08 + P(positive & no diabetes)

P(positive) = 0.08 + 0.092 = 0.172

P(has diabetes | positive) = 0.076/0.172 = 0.442

8 0

3 years ago