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Radda [10]
3 years ago
7

Someone please help me

Mathematics
1 answer:
Grace [21]3 years ago
7 0

ofc i can help! hold up m y mom is vcalling

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7

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6y^-4 <br> simplified<br> please help!!
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1/1296

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6-8x=7x-9x+12<br><br> x = ??
Snezhnost [94]

Answer:

x = -1

Step-by-step explanation:

<u>Step 1:  Combine like terms on the right side </u>

6 - 8x = 7x - 9x + 12

6 - 8x = -2x + 12

<u>Step 2:  Add 8x to both sides </u>

6 - 8x + 8x = -2x + 12 + 8x

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<u>Step 3:  Subtract 12 from both sides </u>

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11 12 c 16 18 if the mean is 12, which number could c be?​
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Step-by-step explanation:

hope this helps

3 0
3 years ago
There are 10 true-false questions and 20 multiple choice questions from which to choose a five-question quiz how many ways can t
beks73 [17]

Answer:

In 68229 ways can the quiz be selected such that there is atleast three multiple choice questions

Step-by-step explanation:

Given:

Number of True or false questions= 10

Number of multiple choice questions= 20

To Find:

How many ways can 5 questions can be selected if there must be at least three multiple choice questions =?

Solution:

Combination

A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.  

The question States there sholud be ATLEAST 3 multiple choice question,

So, we may have

(3 Multiple choice question and 2 true or false question) or

(4 Multiple choice question and 1 true or false question) or

(5 Multiple choice question and 0 true or false question)

Required Number of ways = (20C3 X10C2) +(20C4 X10C1) + (20C5 X10C0)

Required Number of ways =(\frac{20!}{20!(20-3)!}\times\frac{10!}{10!(10-2)!})+(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-1)!}) +(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-0)!})

Required Number of ways = ( 1140 x 42) + (4845 x 10) +(15504 x 1)

Required Number of ways = 47880+48450+15504

Required Number of ways = 68229

3 0
3 years ago
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