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3 years ago

A line passes through the points (0, -8) and (5, 10). What is its equation in slope-intercep

Mathematics

1 answer:

DochEvi [55]3 years ago

5 0

<h2>Answer: y = ¹⁸/₅ x - 8 </h2>

Step-by-step explanation:

For us to write the equation for this line, we need to (1) find the slope of the line, and (2) use one of the points to write an equation:

The question gives us two points, (0, -8) and (5, 10), from which we can find the slope and later the equation of the line.

<u>Finding the Slope</u>

The slope of the line (m) = (y₂ - y₁) ÷ (x₂ - x₁)

= (10 - (- 8)) ÷ (5 - 0)

= ¹⁸/₅

<u>Finding the Equation</u>

We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line:

⇒ y - (-8) = ¹⁸/₅ (x - 0)

we could also transform this into the slope-intercept form ( y = mx + c)

since y - (-8) = ¹⁸/₅ (x - 0)

⇒ y + 8 = ¹⁸/₅ x

∴ y = ¹⁸/₅ x - 8

<em>To test my answer, I have included a Desmos Graph that I graphed using the information provided in the question and my answer.</em>

<em />

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The polynomial X to the 3rd+8 is equal to

NNADVOKAT [17]

Answer:

Step-by-step explanation:

Factor x3+8

x3+8

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3 years ago

As part of an insurance company’s training program, participants learn how to conduct an analysis of clients’ insurability. The

hammer [34]

Answer:

1) cpk < 1.33, therefore it is not capable

b) cpk = 1.33, therefore it is capable

c) cpk < 1.33, therefore it is not capable

2) Cpk can never be greater than the Cp, but can be equal to it

Step-by-step explanation:

Upper limit (USL) = 47 minutes and Lower limit (LSL) = 30 minutes

a) mean (μ) = 37 minutes, standard deviation (σ) = 3 minutes

$cpk=min(\frac{USL-\mu}{3\sigma}, \frac{\mu - LSL}{3\sigma})=min(\frac{47-37}{3*3},\frac{37-30}{3*3} )=min(1.11,0.78)=0.78$

$cp=(\frac{USL-LSL}{6\sigma})=\frac{47-30}{6*3}=0.94$

cpk < 1.33, therefore it is not capable

b) mean (μ) = 38 minutes, standard deviation (σ) = 2 minutes

$cpk=min(\frac{USL-\mu}{3\sigma}, \frac{\mu - LSL}{3\sigma})=min(\frac{47-38}{3*2},\frac{38-30}{3*2} )=min(1.5,1.33)=1.33$

$cp=(\frac{USL-LSL}{6\sigma})=\frac{47-30}{6*2}=1.42$

cpk = 1.33, therefore it is capable

c) a) mean (μ) = 38.5 minutes, standard deviation (σ) = 2.9 minutes

$cpk=min(\frac{USL-\mu}{3\sigma}, \frac{\mu - LSL}{3\sigma})=min(\frac{47-38.5}{3*2.9},\frac{38.5-30}{3*2.9} )=min(0.98,0.98)=0.98$

$cp=(\frac{USL-LSL}{6\sigma})=\frac{47-30}{6*2.9}=0.98$

cpk < 1.33, therefore it is not capable

2) Cpk can never be greater than the Cp, but can be equal to it

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3 years ago

Mary reduced the size of a painting to a width of 3.3 in. What is the new height if it was originally 32.5 in tall and 42.9 in w

posledela

The ratio of the height and width remain constant as you change it's size.
So..
A = Old width
B = Old height
C = New width
D = New height

A/B = C/D

Plug in the values we have

42.9 / 32.5 = 3.3 / D

Multiple both sides by D
D * 42.9 / 32.5 = 3.3

Multiple both sides by 32.5

D * 42.9 = 3.3 * 32.5

Now divide both sides by 42.9

D = 3.3 * 32.5 / 42.9

Now simply do the multiplication and division using the available numbers.
D = 2.5

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