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just olya [345]

2 years ago

5

What is tje equation for the linear function whose graph contains the points (2,5) and (3,0) ?

1 answer:

dolphi86 [110]2 years ago

4 0

In order to find a slope, there is an equation:

Y2-Y1/X2-X1

X1=2
X2=3
Y1=5
Y2=0

0-5= -5
3-2=1

Since slope is written as Y/X, slope would be-5/1 or simply, -5.

Now, we need to find the y-intercept.

To do this, shift the original equation:

Original equation: y=-5x+b

Shifted equation: b=y+5x

Now, plug in one of the points.

Let's use (2,5)

b=5+5(2)

5*2=10
10+5=15

y-intercept=15

So, the full equation of the line would be:

y=-5x+15

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Answer:

The coordinates of point B are (6,7).

Step-by-step explanation:

Given points are

Blue = A(-4,8)

Purple = M(1,7.5)

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Let coordinates of points are (a,b).

Midpoint of two points is

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Coordinates of midpoint are (1,7.5).

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2 years ago

Evaluate C_n.xP^xQn-x For the given n=7, x=2, p=1/2

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Answer:

The value of given expression is $\frac{21}{128}$ .

Step-by-step explanation:

Given information: n=7, x=2, p=1/2

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The given expression is

$C(n,x)p^xq^{n-x}$

It can be written as

$^nC_xp^xq^{n-x}$

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$^7C_2(\frac{1}{2})^2(\frac{1}{2})^{7-2}$

$\frac{7!}{2!(7-2)!}(\frac{1}{2})^2(\frac{1}{2})^{5}$

$\frac{7!}{2!5!}(\frac{1}{2})^{2+5}$

$\frac{7\times 6\times 5!}{2\times 5!}(\frac{1}{2})^{2+5}$

$21(\frac{1}{2})^{7}$

$\frac{21}{128}$

Therefore the value of given expression is $\frac{21}{128}$ .

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