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pshichka [43]
3 years ago
7

Three cards are drawn with replacement from a standard deck. what is the probability that the first card will be a diamond, the

second card will be a black card, and the third card will be a face card
Mathematics
1 answer:
svlad2 [7]3 years ago
6 0
507/17576 or 0.0288 

you take all the probabilities separate then multiply them all together. Then Simplify
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−3 2/7−(−5/7) plz help me
balu736 [363]

Answer:

-2 or -14/7 (they're the same)

Step-by-step explanation:

  1. Write it out: -3\frac{2}{7} - (-\frac{5}{7})  
  2. Simplify. It should now look like this: -3\frac{2}{7} + \frac{5}{7}  
  3. Turn -3 2/7 into an improper fraction. It should now look like this: -19/7
  4. Add: -\frac{19}{7} + \frac{5}{7} = -\frac{14}{7} = -2  

I hope this helps!

3 0
4 years ago
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How do you multiply 1/16 x 16
murzikaleks [220]
16/16 = 1 is the answer to your given question.
8 0
3 years ago
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1092 flowers are arranged into 26 vases, with the same number of flowers in each case.how many flowers would be needed to fill 1
gregori [183]
1092 divided by 28 and you will get 42
And 1092 divided by 130 you will get 8.4
6 0
3 years ago
What is the lateral area of the cone with the raduis of 10 and height 12.1
beks73 [17]

Answer:

493.15

Step-by-step explanation:

3 0
3 years ago
give an example of addition problem in which you would and would not group the addends differently to add
aleksandrvk [35]
An <u>example of a problem</u> where I <em>would not</em> group the addends differently is:

3+2+4.

An <u>example of a problem</u> where I <em>would</em> group the addends differently is:

2+5+8.

Explanation:

In the <u>first problem</u>, I would not group the addends differently before adding.  This is because I cannot make 5 or 10 out of any of the numbers.  We group addends together to make "easier" numbers for us to add, such as 5 and 10.  If we cannot do that, there is no reason to regroup the addends.

In the <u>second problem</u>, I would regroup like this:
2+8+5

This is because 2+8=10, which makes the problem "easier" for us to add.  Since we can get a number like this that shortens the process, we can regroup the addends.
3 0
3 years ago
Read 2 more answers
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