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ankoles [38]
3 years ago
9

Write the following in slope-intercept form.

Mathematics
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

\huge\boxed{y=4x-7}

Step-by-step explanation:

Linear equations will always be in the form y=mx+b, where m is the slope and b is the y-intercept

Since we know nothing about this equation, other than the fact that there are two points in it, we must find the slope and the y-intercept.

Luckily, we have two points to work with. We know that the slope between two points will be the change in y divided by the change in x (\frac{\Delta y}{\Delta x}), so we can use the two points given to us to find both changes.

The y value goes from 1 to 17, which is a 17-1=16 change.

The x value goes from 2 to 6, which is a 6-2=4 change.

Now that we know both changes, we can divide the change in y by the change in x.

\frac{16}{4}=4

Now that we know the slope (4), we can plug it into our equation (y=mx+b).

y=4x+b

Now all we need to do is find the y-intercept. Since we know the slope and one of the points the line passes through, we can find the y-intercept by substituting in the values of x and y. Let's use the point (2, 1).

  • 1 = 4(2) + b
  • 1 = 8+b
  • b = 1-8
  • b = -7

Therefore our y-intercept is -7. Now that we know the slope and the y-intercept, we can plug it into our equation.

y=4x-7

Hope this helped!

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Of all the weld failures in a certain assembly, 85% of them occur in the weld metal itself, and the remaining 15% occur in the b
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Answer:

a.) 0.1028

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c.) 0.0388

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Step-by-step explanation:

Forming a binomial Probability distribution

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Probability of success for base metal failure = 15%

We use the probabilit distribution formula of combination to solve the problem.

P(x=r) = nCr * p^r * q^n-r

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P(x=5) = 20C5 * 0.15^5 * 0.85^15

P(x=5) = 0.1028

b.) probability that fewer than 4 are base metal failure= P(x=0) + P(x=1) + P(x=2) + P(x=3)

P(x=0) = 20C0 * 0.15^0 * 0.85^20 = 0.0388

P(x=1) = 20C1 * 0.15¹ * 0.85^19 = 0.1368

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P(x=3) = 20C3 * 0.15³ * 0.85^17 = 0.2428

Probability that fewer than 4 are base metal failures becomes: 0.038 + 0.1368 + 0.2293 + 0.2428 = 0.6477

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P(x=0) = 0.0388

d.) mean of base metal failures = np = 20*0.15 = 3

e.) standard deviation of base metal failures = √np(1-p)

=3 * (1 - 0.15) = 3 * 0.85

= 2.55

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