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3 years ago

Y+2=-5/4(x-4) what is the slope

1 answer:

4 0

Because Point Slope Form contains m, or slope, in itself, look at the number that is multiplying the parentheses.

-5/4 is m, or slope.

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This is a continuous set of points in the coordinate plane. It can be, but is not always, straight.

castortr0y [4]

Answer:

Linear. While all linear equations produce straight lines when graphed, not all linear equations produce linear functions. In order to be a linear function, a graph must be both linear (a straight line) and a function (matching each x-value to only one y-value).

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2 years ago

What is the x-coordinate of the solution to the following system of equations?

Alja [10]

Answer:

-1,2

Step-by-step explanation:

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2 years ago

Please help with this geometry math problem!

valina [46]

Answer:

∠8 * 4 = ∠7 ----------------(1)

∠8 + ∠7 = 180 ------------(2)

Taking the value of ∠7 from (1) and using it in (2):

∠7 = 4 * ∠8

∠8 + ∠7 = 180

∠8 + (4 * ∠8) = 180

∠8 + 4∠8 = 180

5∠8 = 180

∠8 = 36° ---------------(3)

Using the value of ∠8 from (3) in (2):

∠8 + ∠7 = 180

36 + ∠7 = 180

∠7 = 144° and ∠8 = 36°

4 0

3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$