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zepelin [54]
3 years ago
13

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE I BEGGING YOUPLZ

HELP plz show your work ( 77 POINTS)
1. 1/4 ÷ 2 1/4


2. Solve and graph the solution to each inequality below.

2(3y-2)>8


3. Solve and graph the solution to the inequality below.

2(k-1)>-2.2


4. Solve and graph the solution to the inequality below.

4x+24≤ 32


5. Solve and graph the solution to the inequality below. 5n+5>-45
Mathematics
1 answer:
MAXImum [283]3 years ago
6 0

Answer 1 -16 3y-2 >16 k-1 >24 -8.8x>24-8.8x<32

Step-by-step explanation:

k>3, x> - 10-11,y>2-3 (128k-127)

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write the ratio for the following description. kaleel made three times as many baskets as john during basketball practice.
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Ratio is 3:1


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5 0
3 years ago
-6x-7=3x+11 how to solve this problem
Shalnov [3]
You have to get the x to one side so you'll subtract 3x from -6x and 3x from 3x, the 3x will cancel out and -6x - 3x is -9x-7=11, you have to add 7 from both sides the 7+7 cancels out  and also add 11+7 which leaves -9x=18, you have to divide -9x/-9 and do 18/-9, which gets you x=-2. Here's the work shown if you didn't what i said...
-6x-7=3x+11
-3x    -3x
-9x-7=11
    +7  +7
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6 0
2 years ago
Middle School: Math (10 Points)
sasho [114]
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:

A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?

You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked

2) a - 2%  as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.  

b - 1,000,000/2500 = 400
400 packages are defective each year
5 0
3 years ago
What is the sum: (1−5q)+2(2.5q+8)
mash [69]

The answer is seventeen

7 0
2 years ago
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