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4 years ago

Write an equation in slope-intercept form of the line parallel to y=8x-8 through (5,2)

Mathematics

1 answer:

AlekseyPX4 years ago

3 0

Answer:

The equation would be y = 8x - 38

Step-by-step explanation:

To find the equation of this new line, we first need to identify the slope of the first line. In slope-intercept form, the slope is always the coefficient of x. This makes the slope of the first equation 8. Since parallel lines have the same slope, our new slope will also be 8.

Given that and the point, we can solve for the equation using point-slope form.

y - y1 = m(x - x1)

y - 2 = 8(x - 5)

y - 2 = 8x - 40

y = 8x - 38

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. A condo in Orange Beach, Alabama, listed for $1.4 million with 20% down and financing at 5% for 30 years. What is the monthly

vovikov84 [41]

Listed price = $1.4 million
Down payment = 20% of $1.4 million = 0.2 x 1,400,000 = 280,000
Amount left to pay = $1.4 million - 280,000 = $1,120,000
Present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt) / r
where: PV = $1,120,000
r = 5% = 0.05
t = 12
n = 30 years.
1,120,000 = P(1 - (1 + 0.05/12)^-(12 x 30)) / 0.05
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3 years ago

A basketball player signs a contract that pays him 16 million over 4 years. What is his average pay per year?

nirvana33 [79]

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3 years ago

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Using the grid box, draw a net for a rectangular prism with a length of 8 units, width of 2 units and height of 3 units

Alexandra [31]

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3 years ago

Which is bigger 9% or 0.4

gulaghasi [49]

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3 years ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

3 years ago