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Fantom [35]
3 years ago
7

Point C is located at (1, 2) and point D is located at (−4, −2). Find the x value for the point that is 1\4 the distance from po

int C to point D
Mathematics
1 answer:
max2010maxim [7]3 years ago
7 0

The easiest way to do this is to realize this is a <em>weighted average</em> of the two points.  (A weighted average is a linear combination where the coefficients add to one.)


(x,y) = (1-t)C + tD


where t is a real parameter. When t=0 we're at point C, when t=1 we're at point D. We're interested in t=1/4,


(x,y)= (3/4) C + (1/4) D


We're only asked for the x coordinate:


x = (3/4) (1) + (1/4) (-4) = 3/4 - 1 = - 1/4


Answer: \quad - \dfrac 1 4



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kodGreya [7K]

Answer:

(-∞,7) U (7,∞)

Step-by-step explanation:

f(x)= x+2

g(x) = x-7

\frac{f(x)}{g(x)} =\frac{x+2}{x-7}

Here we have x-7 in the denominator

To find domain we set the denominator =0  and solve for x

x-7=0

Add 7 on both sides

x=7

x=7 makes the denominator 0 that is undefined

So we ignore 7 for x

Hence domain is

(-∞,7) U (7,∞)

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3 years ago
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397 students went on a field trip. Eight
lbvjy [14]

Answer:48

Step-by-step explanation:

Total number of students=397

Students that travelled in a car=29

Let the students that travelled in a bus be represented by y.

Since 8 buses were filled, it will be 8y for total students

8y + 29= 397

8y= 397-29

8y= 368

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Simplify 2y+(4+5y) expression
Pavlova-9 [17]
2y+(4+5y)
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Which statement is true?
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Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam
Vlad [161]

Answer

P(A) = 0.30

P(B) = 0.77

P(A\ n\ B) = 0.22

P(A\ u\ B) = 0.85

Explanation:

Given

See attachment for proper data presentation

n = 100 --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

Here:

Yes = 22 and No = 8

n(A) = Yes + No

n(A) = 22 + 8

n(A) = 30

P(A) is then calculated as:

P(A) = \frac{n(A)}{Sample}

P(A) = \frac{30}{100}

P(A) = 0.30

Solving (b): P(B)

We only consider data in the Yes column.

Here:

(1) = 22    (2) = 25 and (3) = 30

n(B) = (1) + (2) + (3)

n(B) = 22 + 25 + 30

n(B) = 77

P(B) is then calculated as:

P(B) = \frac{n(B)}{Sample}

P(B) = \frac{77}{100}

P(B) = 0.77

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

i.e. [Supplier 1][Yes]

This is represented as: n(A n B)

n(A\ n\ B) = 22

The probability is then calculated as:

P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}

P(A\ n\ B) = \frac{22}{100}

P(A\ n\ B) = 0.22

Solving (d): P(A u B)

This is calculated as:

P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)

This gives:

P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}

Take LCM

P(A\ u\ B) = \frac{30+77-22}{100}

P(A\ u\ B) = \frac{85}{100}

P(A\ u\ B) = 0.85

7 0
3 years ago
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