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Radda [10]
3 years ago
7

Divide the number by 3 and adds 2

Mathematics
1 answer:
anzhelika [568]3 years ago
8 0

aiejrhejekwhvqiekwnww

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A bag contains 5 blue marbles, 6 red marbles, and 9 green marbles. Two marbles are drawn at random, one at a time and without re
ArbitrLikvidat [17]
Should be C. 1/19. because if there's 20 maebles in the bag and without replacing the marble then it actually would be 2/18
5 0
3 years ago
Ashley's mother tells her to see 3/5 of her dollar how much is that
oksian1 [2.3K]
Your answer will be D 0.60

A dollar is 1.00 so 0.60 will be 3/5
5 0
3 years ago
What is 6,383 in expended form?<br>​
liberstina [14]

Answer:

expanded notation form:

 6,383 =

 6,000  

+ 300  

+ 80  

+ 3  

expanded factors form:

   6,383 =

 6 × 1,000  

+ 3 × 100  

+ 8 × 10  

+ 3 × 1  

expanded exponential form:

 6,383 =

6 × 103

+ 3 × 102

+ 8 × 101

+ 3 × 100

Step-by-step explanation:

i wasn't entirely sure what you needed but i hope this helps :) have a great day !!

8 0
3 years ago
Calculate the average travel time for each distance, and then use the results to calculate. A 6-column table with 3 rows in the
BigorU [14]

The average travel time is simply the mean travel time between the given points

  • The average time that it takes for the car to travel the first 0.25m is 2.23s.
  • The average time to travel just between 0.25m and 0.50m is 0.9s.
  • Given the time taken to travel the second 0.25 m section, the velocity would be 0.28m/s.

<u>(a) The average time to travel the first 0.25m</u>

The time travel in the first 0.25m are: 2.24s, 2.21s and 2.23s.

So, the average time to travel is:

Time = \frac{2.24s + 2.21s + 2.23s}{3}

Time = \frac{6.68s}{3}

Time = 2.23s

<u>(b) The average time to travel just between 0.25m and 0.50m</u>

The time travel in the 0.50m are: 3.16s, 3.08s and 3.15s.

So, the average time to travel this distance is:

Time = \frac{3.16s + 3.08s + 3.15s}{3}

Time = \frac{9.39s}{3}

Time = 3.13s

The average time to travel between both distance is the difference between the average time of each distance.

So, we have:

Average = 3.13s - 2.23s

Average = 0.9s

<u>(c) The velocity in the second 0.25m section</u>

The distance and time are:

Distance = 0.25m

Time = 0.9s

So, the velocity is:

Velocity = \frac{Distance}{Time}

This gives

Velocity = \frac{0.25m}{0.9s}

Velocity = 0.28m/s

Read more about distance, velocity and time at:

brainly.com/question/4931057

6 0
3 years ago
HELP PLEASE PLEASE :’(
Svet_ta [14]

Practice:

1. A.

2. D. y=5+0.84x-0.03x^2, x>=0

3. B. y=2/x, x>0

Quiz:

1. A C E. (-4,4) (0,0) (6,9)

2. A. y=-1/3x+19/3

3. D. x=2y^2-1, y>=0

4. B. y=3/x^2, x>0

8 0
3 years ago
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