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3 years ago

Kiana wants to write equations in the form y = mx + b for the lines passing through

Mathematics

1 answer:

horrorfan [7]3 years ago

3 0

Answer:

'Substitute the slope and the coordinates of point P(-3,2) in y = mx + b and then solve for b in each equation'.

Step-by-step explanation:

Kiana wants to write equations in the form y = mx + b for the lines passing through point P that are parallel and perpendicular to line q.

This equation is in the slope-intercept form where m is the slope and b is the y-intercept.

First, she finds the slope of line q and the slope of line s to be – 2. Point P is located at (-3,2).

Therefore, the step that can be used to find the y-intercept is 'substitute the slope and the coordinates of point P(-3,2) in y = mx + b and then solve for b in each equation'. (Answer)

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Look at this graph: What are the coordinates of the vertex?

asambeis [7]

Answer:

-1, -1

Step-by-step explanation:

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3 years ago

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A business buys a computer for $3,000. After 4 years the value of the computer is expected to be $250. The value, v, can be rela

yuradex [85]

You are given two points in the linear function. At time 0 years, the value is $3000. At time 4 years, the value is $250. This means you have points (0, 3000) and (4, 250). You need to find the equation of the line that passes through those two points.

y = mx + b

m = (y2 - y1)/(x2 - x1) = (3000 - 250)/(0 - 4) = 2750/(-4) = -687.5

Use point (0, 3000).

3000 = -687.5(0) + b

b = 3000

The equation is

y = -687.5x + 3000

Since we are using points (t, v) instead of (x, y), we have:

v = -687.5t + 3000

Answer: d. v = -687.50 t + 3,000

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3 years ago

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$