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3 years ago

What is the length of segment DA?

Mathematics

2 answers:

Svet_ta [14]3 years ago

8 0

Answer: DA = 6

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Explanation:

The triangles are similar, allowing us to make the proportion below

DB/DA = DA/DC

Plug in the given values. Cross multiply and solve for x

DB/DA = DA/DC

9/DA = DA/4

9*4 = (DA)*(DA)

36 = (DA)^2

(DA)^2 = 36

DA = sqrt(36)

DA = 6

The length of DA is exactly the geometric mean of the lengths BD and DC

geometric mean of x and y = sqrt(x*y)

frozen [14]3 years ago

6 0

Answer:

Step-by-step explanation:

The length of this line segment is the distance between its endpoints A and B. So, a line segment is a piece or part of a line having two endpoints. Unlike a line, a line segment has a definite length.

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Describe a real world situation that can be modeled by 5t

jeka57 [31]

I used to shovel snow for $5 an hour.

If ' t ' was the number of hours I spent shoveling somebody's driveway,
then ' 5t ' was the number of dollars he owed me for the job.

' t ' didn't even have to be a whole number. It worked fine with any number.

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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

This was on my homework 5 X-5 = 5x+7

KengaRu [80]

Answer:

No solution

Step-by-step explanation:

Original Equation:

5x - 5 = 5x + 7

Add 5 to both sides

5x = 5x + 12

So we already know that this equation is not true, making it no solution

Hope this helped

6 0

3 years ago

I need to know the steps on how to do it

Sergeu [11.5K]

You need help with what problem

4 0

4 years ago

Please help. I’ll mark you as brainliest if correct!

sesenic [268]

Answer:

Quantity (lbs) of type 1 candy x = 8

Quantity (lbs) of type 2 candy y = 17,5

Step-by-step explanation:

Let´s call "x" quantity (in pounds) of candy type 1 in the mixture, and "y" quantity (in pounds ) of candy type 2, then according to the problem statement.

x + y = 25,5

2,20*x + 7,30*y = 5,70 * 25,5 ⇒ 2,20*x + 7,30*y = 145,35

Then we have a two equation system

x + y = 25,5 ⇒ y = 25,5 - x

2,20*x + 7,30*y = 145,35 ⇒ 2,20*x + 7,30* (25,5 - x ) = 145,35

2,20*x + 186,15 - 7,30*x = 145,35

5,1*x = 40,8

x = 40,8/5,1

x = 8 lbs

And y = 25,5 - 8

y = 17,5 lbs

5 0

3 years ago