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2 years ago

Write an equation of the line that passes through (18, 2) and is parallel to the line 3y−x=−12

Mathematics

2 answers:

Flauer [41]2 years ago

6 0

Answer:

$y = \frac{x}{3} - 4$

I hope this helps!

Stolb23 [73]2 years ago

5 0

Answer:

y= -4x+74

Step-by-step explanation:

hope that helped

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Please help! I'm having problems with deciding if 2.80 would be 2 or less or 2 or more.

Gekata [30.6K]

2.80 is more than 2, i would round it.

So sorry if this is wrong :(

7 0

3 years ago

2. Suppose over several years of offering AP Statistics, a high school finds that final exam scores are normally distributed wit

nirvana33 [79]

Answer:

By the Central Limit Theorem, the mean is 78, the standard deviation is $s = \frac{6}{\sqrt{n}}$ and the shape is approximately normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 78 and a standard deviation of 6

This means that $\mu = 78, \sigma = 6$

Samples of n:

This means that the standard deviation is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{n}}$

What are the mean, standard deviation, and shape of the distribution of x-bar for n?

By the Central Limit Theorem, the mean is 78, the standard deviation is $s = \frac{6}{\sqrt{n}}$ and the shape is approximately normal.

6 0

3 years ago

Answer ASAP plz I have 2 minutes

earnstyle [38]

250+25= 275 is your answer

5 0

3 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago

I forget how to do this problem, little help-?

jenyasd209 [6]

X=4

Please mark brainliest

4 0

3 years ago