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Aneli [31]
2 years ago
10

Are the two sets equivalent? C={4,6,3,2,1}, D={3,2,6,1,4}.

Mathematics
1 answer:
slavikrds [6]2 years ago
4 0

Answer:

yes they are

Step-by-step explanation:

tho the numbers are in different order

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Use the order of operations and the digits 2, 4, 6, and 8 to create an expression with a value of 2
Naya [18.7K]
4÷(6-(8÷2))=2

Do the order of operations, and the end value will be 2.

Do you understand how I got that? Good luck by the way!
3 0
3 years ago
Pls help meh<br> i will give brainlyest
MissTica

Answer:

The answer is 75

can me be brainliest?

Step-by-step explanation:

PEMDAS

20x4-100/20

80-5

75

5 0
3 years ago
Read 2 more answers
SIMPLE MATH PROBLEM PLZ ANSWER<br><br> u can pick more than one..
7nadin3 [17]

its the first one 6k - 5

if u were to distribute u need to remember that 6 is negative so it changes -k to a positive 6k and then you would time 4 by -6 which would be -24 but u would just write it like 19 + 6k - 24. then u would add 19 to both sides by 19 which would be like adding 19 to -24 which would equal to -5 so it would equal to 6k - 5.

hope I explained the steps well enough for you.

5 0
3 years ago
Ví dụ 1.5. Một người đi mua hàng 3 lần. Xác suất lần đầu mua được hàng tốt là 0,7.
vodka [1.7K]

Answer:

a) 0.50575,

b) 0.042

Step-by-step explanation:

Example 1.5. A person goes shopping 3 times. The probability of buying a good product for the first time is 0.7.

If the first time you can buy good products, the next time you can buy good products is 0.85;  (I interpret this as, if you buy a good product, then the next time you buy a good product is 0.85).

And if the last time I bought a bad product, the next time I bought a good one is  0.6. Calculate the probability that:

a) All three times the person bought good goods.

P(Good on 1st shopping event AND Good on 2nd shopping event AND Good on 3rd shopping event) =

P(Good on 1st shopping event) *P(Good on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st and 2nd shopping events yield Good) =

(0.7)(0.85)(0.85) =

0.50575      

b) Only the second time that person buys a bad product.

P(Good on 1st shopping event AND Bad on 2nd shopping event AND Good on 3rd shopping event) =

P(Good on 1st shopping event) *P(Bad on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st is Good and 2nd is Bad shopping events) =

(0.7)(1-0.85)(1-0.6) =

(0.7)(0.15)(0.4) =

0.042

5 0
2 years ago
Help pls<br> Factor out the GCF (greatest common factor).<br> 16x^2-12x^4
Vika [28.1K]

Answer:

It is 4

Step-by-step explanation:

I really hope I helped you Have a nice day

3 0
3 years ago
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