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Ugo [173]
3 years ago
5

Can anyone help me with these two questions

Mathematics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

Step-by-step explanation:

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What’s the calculation of the equation 5/8-1/4=
Kay [80]

Answer: 3/8

Explanation: To subtract unlike fractions such as 5/8 - 1/4, first find a common denominator. The common denominator of 8 and 4 will be the least common multiple of 8 and 4 which is 8.

Notice that our first fraction already has 8 in the denominator so it stays the same. To get 8 in the denominator of 1/4, we multiply the numerator and denominator by 2 to get 2/8.

Now we have 5/8 - 2/8.

To subtract like fractions, we simply subtract the numerators to get 3 and keep our denominator of 8 and we have 3/8.

Since 3/8 is in lowest terms, it's our final answer.

Therefore, 5/8 - 1/4 = 3/8.

4 0
3 years ago
8x^2-x^3<br> Help me I need the answer
mario62 [17]

Answer:

-x^2(x-8)

Step-by-step explanation

7 0
3 years ago
Is (3,1) a solution of the equation y=-x+3
kogti [31]

no i don't think so the answer will be (3,0)

3 0
3 years ago
Read 2 more answers
Please help me figure out the right answer and if you don’t know please don’t answer the question
Degger [83]

Answer:

Figure A and C

Step-by-step explanation:

3 0
3 years ago
Q3. Consider two identical pack of cards A and B. When one card from A is taken and shuffled with the card B, the first top card
vivado [14]

Answer:

0.02

Step-by-step explanation:

Pack A and B both consisted of 52 cards. When shuffled, 51 cards in A and 53 cards in deck B. It appears that the Top card of A left with only 51 possible cards to be taken from A (since we know the Queen hadn't moved).

The two possible cases are: probability of not taking King of Hearts (from A to B), and the other were we did.

Let K= the event of taking the King of Hearts from A to B and be B_{K} the occasion being the top card of B.

P(B_{K} )=P(K)P(B_{K} |K)+P(K^{C} )P(B_{K} |K^{C} )\\

=\frac{1}{51} \frac{2}{53} +\frac{50}{51} \frac{1}{53}

=\frac{52}{2703} \\=0.02

8 0
3 years ago
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