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Vilka [71]
2 years ago
6

f the coordinates of point A are (8 , 0) and the coordinates of point B are (3 , 7), the y-intercept of is . If the coordinates

of point D are (5 , 5), the equation of line is y = x + .
Mathematics
1 answer:
Doss [256]2 years ago
8 0

The equation of the line is y = -7/5x + 56/5

<h3>Equation of a line</h3>

The formula for expressing equation of a line is given as y = mx + b

where

m is the slope

b is the y-intercept

Given the coordinates A(8, 0) and (3, 7)

Determine the slope

Slope =7-0/3-8

Slope = -7/5

For the intercept

0 = -7/5(8) + b

b = 56/5

Hence the equation of the line is y = -7/5x + 56/5

Learn more on equation of a line here:brainly.com/question/22119282

#SPJ1

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Why is a rhombus not a rectangle?
alukav5142 [94]

Answer:

A rectangle is a parallelogram with all its interior angles being 90 degrees. A rhombus is a parallelogram with all its sides equal. This means that for a rectangle to be a rhombus, its sides must be equal.



Hope this helps! Please mark brainliest! Have a great day!


Thank you for using Brainly!

4 0
3 years ago
To solve -8p = 48, which of the following could you do to both sides of the equation? A. add -8 B. subtract -8 C. multiply by -8
masya89 [10]

Answer:

D. divide by -8

Step-by-step explanation:

8 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
HELP I NEED HELP ASAP
Monica [59]

Answer:

A

Step-by-step explanation:

3 0
3 years ago
"John and two friends rent a canoe at a park. Each person much rent a life jacket. If the bill for the rental of the canoe and l
Slav-nsk [51]
I would need to know the amount of $ the rental life jackets and canoe are.
6 0
3 years ago
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