Input in standard form the equation of the given line.

1 answer:

5 0

Use equation of line which passes through the 2 points $A(x_1,y_1),B(x_2,y_2)$ .

$y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1) \\ \\A(1,5),B(-2,3)
\\ y-5= \frac{3-5}{-2-1}(x-1)
\\y-5= \frac{2}{3} (x-1)
\\y-5=\frac{2}{3}x-\frac{2}{3}
\\3y-15=2x-2
\\-2x+3y=-2+15
\\-2x+3y=13$

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Given the system of equations: 2x – y = –2 x = 14 2y

SSSSS [86.1K]

If you would like to solve the system of equations, you can do this using the following steps:

2x - y = -2
x = 14 + 2y
__________
2x - y = -2
2 * (14 + 2y) - y = -2
28 + 4y - y = -2
28 + 3y = -2
3y = -2 - 28
3y = -30 /3
y = -30/3
y = -10

x = 14 + 2y = 14 + 2 * (-10) = 14 - 20 = -6

(x, y) = (-6, -10)

The correct result would be (-6, -10).

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2 years ago

A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is kno

Reika [66]

Answer:

The 90th percentile of the distribution is 6.512 ml.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 6 milliliters (ml) and a standard deviation of 0.4 ml.

This means that $\mu = 6, \sigma = 0.4$

Find the dye amount that represents the 90th percentile (i.e. 90%) of the distribution.

This is X when Z has a p-value of 0.9, so X when Z = 1.28. Then

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 6}{0.4}$

$X - 6 = 1.28*0.4$

$X = 6.512$

The 90th percentile of the distribution is 6.512 ml.

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2 years ago

Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, x = 2, and y = 1 about the y-axis.

andrezito [222]

Since the rotation is about the y-axis, I'll integrate by dy.

$\displaystyle y=x^3\\x=\sqrt[3]y\\\\V=\pi \int \limits_1^8(2^2-(\sqrt[3]y)^2)\, dy\\V=\pi \Big[4x-\dfrac{3}{5}x^{\tfrac{5}{3}}\Big]_1^8\\V=\pi \left(4\cdot8-\dfrac{3}{5}\cdot8^{\tfrac{5}{3}\right-\left(4\cdot1-\dfrac{3}{5}\cdot1^{\tfrac{5}{3}\right)\right)\\V=\pi \left(32-\dfrac{96}{5}-\left(4-\dfrac{3}{5}\right)\right)\\V=\pi \left(\dfrac{64}{5}-\dfrac{17}{5}\right)\\V=\dfrac{47\pi}{5}$