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Alina [70]
3 years ago
13

Find the slope of the line that passes through (1,6) and (4,4)

Mathematics
1 answer:
Vadim26 [7]3 years ago
8 0

Answer:

y=-2/3x+20/3

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(4-6)/(4-1)

m=-2/3

y-y1=m(x-x1)

y-6=-2/3(x-1)

y=-2/3x+2/3+6

y=-2/3x+2/3+18/3

y=-2/3x+20/3

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A teacher wants to see if a new unit on factoring is helping students learn. She has five randomly selected students take a pre-
sattari [20]

Answer:

Step-by-step explanation:

Given the sample data

Pre-test... 12 14 11 12 13

Post-Test 15 17 11 13 12

The mean of pre-test

x = ΣX / n

x = (12+14+11+12+13) / 5

x = 12.4

The standard deviation of pre-test

S.D = √Σ(X-x)² / n

S.D = √[(12-12.4)²+(14-12.4)²+(11-12.4)²+(12-12.4)²+(13-12.4)² / 5]

S.D = √(5.2 / 5)

S.D = 1.02.

The mean of post-test

x' = ΣX / n

x' = (15+17+11+13+12) / 5

x' = 13.6

The standard deviation of post-test

S.D' = √Σ(X-x)² / n

S.D' = √[(15-13.6)²+(17-13.6)²+(11-13.6)²+(13-13.6)²+(12-13.6)² / 5]

S.D = √(23.2 / 5)

S.D = 2.15

Test value

t = (sample difference − hypothesized difference) / standard error of the difference

t = [(x-x') - (μ- μ')] / (S.D / n — S.D'/n)

t = (12.4-13.6) - (μ-μ')/ (1.02/5 - 2.15/5)

-1.5 = -1.2 - (μ-μ') / -0.226

-1.5 × -0.226 = -1.2 -(μ-μ')

​0.339 = -1.2 - (μ-μ')

(μ-μ') = -1.2 -0.339

μ-μ' = -1.539

Then, μ ≠ μ'

We can calculate our P-value using table.

This is a two-sided test, so the P-value is the combined area in both scores.

The p-value is 0.172

The p value > 0.1

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2 years ago
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Ede4ka [16]
The answer is A

EXPLANATION:

4 0
3 years ago
2. Given the equation of a circle in standard form, identify the center, radius, and graph the circle.
Varvara68 [4.7K]

Answer:

Center: ( -1 , 2 )

Radius: 6

Step-by-step explanation:

The equation for a circle is given as follow:

(x-h)^{2} +(y-k)^{2} =r^{2}

Where,

the Center is: ( h , k ) (note that the signs of the number are different)

and the radius is: r

So if we compare the original circle equation to the equation in the question we can see that:

(x+1)^{2} +(y-2)^{2} =36

the Center is: (-1,2)

and the radius is: \sqrt{36} = 6

2. To draw the graph find points that lay on the circle, it's better to take the values of x and y from the Center:

first sub y=2 in the equation to find the values for x:

(x+1)^{2} +(y-2)^{2} =36

(x+1)^{2} +(2-2)^{2} =36

(x+1)^{2} +(0)^{2} =36

(x+1)^{2}  =36

x+1 =±\sqrt{36}

x=6-1    AND    x=-6-1

x=5          AND    x=-7

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second sub x= -1 in the equation to find the values for y:

(x+1)^{2} +(y-2)^{2} =36

(-1+1)^{2} +(y-2)^{2} =36

(0)^{2} +(y-2)^{2} =36

(y-2)^{2} =36

y-2=±\sqrt{36}

y=6+2    AND    y=-6+2

y=8          AND    y=-4

  • The points are D(-1,8)  and  E(-1,-4)      

After finding the points write them in the graph and match them together to get the like the circle in the picture below:

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Answer:

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Step-by-step explanation:

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