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2 years ago

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Write the standard form of the line that contains a slope of and y-intercept of 3. Include your work in your final answer. Type

your answer in the box provided or use the upload option to submit your solution.

2 answers:

coldgirl [10]2 years ago

7 0

Y=mx + b
m= slope
b= y intercept

y=3x + 3

mash [69]2 years ago

7 0

Keywords:

<em>standard equation, straight line, slope, cut point </em>

For this case, we must write the equation of a line in a standard way. Which is given by: y = mx + b. Where "m" is the slope and "b" is the cut point with the y axis. The slope is given by:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

To find the slope, substitute two points $(x_ {1}, y_ {1})\ and\ (x_ {2}, y_ {2})$ through which the line passes.

In this case, we were given the cut point $b = 3$ , so we have the following equation:

$y = mx + 3$

Answer:

$y = mx + 3$

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If f (x) = x + 7 and g (x) = StartFraction 1 Over x minus 13 EndFraction, what is the domain of (f circle g) (x)? StartSet x ver

nekit [7.7K]

Answer:

The domain of the function will be {x I x ≠ 13}.

Step-by-step explanation:

The two functions f(x) and g(x) are given to be

f(x) = x + 7 and

$g(x) = \frac{1}{x - 13}$

Now, we have to find the composite function (fog)(x).

Here, (fog)(x) = f{g(x)}

⇒ $(fog)(x) = f(\frac{1}{x - 13}) = \frac{1}{x - 13} + 7 = \frac{7x - 90}{x - 13}$

Therefore, the denominator of the function can not be zero and the domain of the function will be {x I x ≠ 13}. (Answer)

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2 years ago

Read 2 more answers

A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.

Dennis_Churaev [7]

Answer:

Margin of error at 90% is 0.024

Margin of error at 99% is 0.037

Step-by-step explanation:

Sample size = 1076

A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.

So, No. of adults said they believe ghosts = $\frac{64}{100} \times 1076=688.64\sim 688$

So, x = 688

n = 1076

$\widehat{p} = \frac{x}{n}$

$\widehat{p} = \frac{688}{1076}$

$\widehat{p} = 0.639$

$ME=z \times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

z at 90% confidence is 1.64

$ME=1.64 \times \sqrt{\frac{0.639(1-0.639)}{1076}}$

$ME=0.024$

So, margin of error at 90% is 0.024

Find the margin of error needed to be 99% confident.

z at 99% confidence is 2.58

$ME=2.58 \times \sqrt{\frac{0.639(1-0.639)}{1076}}$

$ME=0.024$

So, margin of error at 99% is 0.037

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2 years ago

A plant can either have a smooth seed (p) or a rough seed (q). Rough seeds are recessive and smooth seeds are dominant. In a pop

adell [148]

To calculate this, the Hardy-Weinberg principle can be used:

p² + 2pq + q² = 1 and p + q = 1

where p and q are the frequencies of the alleles (p - dominant, q - recessive), and p², q² and 2pq are the frequencies of the genotypes.

a) Since 32 plants have rough seed (recessive genotype: q²) out of 100 plants in total, then

q² = 32/100 = 0.32

b) q = √q² = √0.32 = 0.56

c) Since p + q = 1, then

p = 1 - q = 1 - 0.56 = 0.44

d) 19 plants with rough seeds (recessive genotype: q²) in a population of 100 means that q² = 19/100 = 0.19

We need to calculate p (the allele frequency for smooth seeds).
We can find q because we know q²:

q = √q² = √0.19 = 0.44

Since p + q = 1, then

p = 1 - q = 1 - 0.4 = 0.56

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2 years ago

Moira borrowed $4,500 from her grandfather to pay for her first year of college. Three years later, she repaid the $4,500 along

Nataly [62]

Answer: 1.8%

Step-by-step explanation:

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2 years ago

A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every ba

Ahat [919]

Answer:

$z=\frac{0.21 -0.15}{\sqrt{\frac{0.15(1-0.15)}{100}}}=1.68$

Step-by-step explanation:

Information provided

n=100 represent the random sample taken

X=21 represent the number of bags overfilled

$\hat p=\frac{21}{100}=0.21$ estimated proportion of overfilled bags

$p_o=0.15$ is the value that we want to test

z would represent the statistic

Hypothesis

We need to conduct a hypothesis in order to test if the true proportion of overfilled bags is higher than 0.15.:

Null hypothesis: $p =0.7$

Alternative hypothesis: $p > 0.15$

The statistic for this case is:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

And replacing the info given we got:

$z=\frac{0.21 -0.15}{\sqrt{\frac{0.15(1-0.15)}{100}}}=1.68$

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