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Irina-Kira [14]
3 years ago
15

What is an equation of the line that passes through point (5,-8) and is parallel to the line x+y=7

Mathematics
1 answer:
Igoryamba3 years ago
4 0

Answer:

The line given is equal to y=-x+7 (Through rearranging)

To find the new line, we substitute the given coordinates (5,8) in the equation without the given y-intersect (+7):

-8=-5+c

c=-3

The line parallel to x+y=7 that passes through the point (5,-8) is y=-x-3

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3 years ago
Solve for f: 6f + 9g = 3g + f
lozanna [386]
Hi there!

Let's solve this equation step by step!
6f + 9g = 3g + f

To solve for f, we need to bring all the terms in the equation with an f in it to the left, and all the other terms (with a g) to the right.

First subtract f from both sides.
5f + 9g = 3g

Now subtract 9g from both sides.
5f = -6g

And finally divide both sides by 5.
f = (-6/5)g

Hence, your answer;
f =  -  \frac{6}{5} g

~ Hope this helps you!
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3 years ago
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Sonja [21]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the
ziro4ka [17]

Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0
4 years ago
Consider the graphed function. 100POINTS!!!!
Sati [7]

Answer:

Domain -5 ≤x<1

Range -4 ≤y<7

Step-by-step explanation:

The domain is the values that x takes

X goes from -5 included to 1 not included

-5 ≤x<1

The range is the values that y takes

y goes from -4 included to 7  not included

-4 ≤y<7

4 0
3 years ago
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