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natulia [17]
2 years ago
6

what is the equation of a line that passes through the point (1,3) and is parallel to a line with a slope of -2?

Mathematics
1 answer:
natta225 [31]2 years ago
3 0

Answer:

Step-by-step explanation:

y - 3 = -2(x - 1)

y - 3 = -2x + 2

y = -2x + 5

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Rama09 [41]

Answer:

x= 17/5

Step-by-step explanation:

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2 years ago
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Can someone please help me understand how to answer for x, y? Thank you so much!
kodGreya [7K]

Answer:

X = 45°

Y = 72°

Step-by-step explanation:

There is a supplementary angle (an pair of angles that create a straight line is equal to 180°) and there is the rule that all three angles in a triangle equal 180° as well.

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2 years ago
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A marketing researcher wants to find a 96% confidence interval for the mean amount those visitors spend per person per day while
ki77a [65]

Answer:

n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=12 represent the population standard deviation

n represent the sample size  

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is \alpha=1-0.96 =0.04. And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got z_{\alpha/2}=2.054, replacing into formula (b) we got:

n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38

So then the minimum sample to ensure the condition given is n= 38

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Radda [10]

Two main facts are needed here:

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2. The square root \sqrt x exists for x\ge0.

(in both cases we're assuming real-valued functions only)

By (2) we know that \sqrt{2x-1} exists if 2x-1\ge0, or x\ge\dfrac12.

By (1), we know that \log(\sqrt{2x-1}+3) exists if \sqrt{2x-1}+3>0, or \sqrt{2x-1}>-3. But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require x\ge\dfrac12, so the function has domain \left[\dfrac12,\infty).

To determine the range, we need to know that, in their respective domains, \sqrt x and \log x increase monotonically without bound. We also know that x=\dfrac12 at minimum, at which point the square root term vanishes, so the least value the function takes on is \log3. Then its range would be [\log3,\infty).

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3 years ago
Can some one explain it 2 me i will appreciate it!
hichkok12 [17]

Answer: y = 6x - 4

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