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Sveta_85 [38]
3 years ago
13

PLEASE HELP ASAP!!!!!!

Mathematics
1 answer:
Oksanka [162]3 years ago
4 0
B cause it makes sense
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A bag contains yellow marbles and blue marbles, 73 in total. The number of yellow marbles is 7 more than 5 times the number of b
Roman55 [17]

Answer:

number of yellow marbles=62.

Step-by-step explanation:

It is given that in a bag there are yellow marbles and blue marbles and the total number of marbles=73.

Let the number of Blue marbles=x, then according to the question, number of yellow marbles will be =5x+7.

Now, Total number of marbles= number of yellow marbles+number of blue marbles.

⇒73=x+5x+7

⇒73-7=6x

⇒66=6x

⇒x=11

Therefore, numberof blue marbles=x=11 and

number of yellow marbles=5x+7=5(11)+7=55+7=62.

8 0
4 years ago
4√6 •√3 how do I show work for this because the answer is 2√12​
Yuliya22 [10]

Answer:

4 \sqrt{6}  \times  \sqrt{3}  = 12 \sqrt{2}

Step-by-step explanation:

We want to simplify the radical expression:

4 \sqrt{6}  \times  \sqrt{3}

We write √6 as √(2*3).

This implies that:

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2 \times 3}   \times  \sqrt{3}

We now split the radical for √(2*3) to get:

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2}  \times  \sqrt{3}  \times  \sqrt{3}

We obtain a perfect square at the far right.

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2}  \times  (\sqrt{3} )^{2}

This simplifies to

4 \sqrt{6}  \times  \sqrt{3}  = 4 \sqrt{2}  \times 3

This gives us:

4 \sqrt{6}  \times  \sqrt{3}  = 4 \times 3 \sqrt{2}

and finally, we have:

4 \sqrt{6}  \times  \sqrt{3}  = 12 \sqrt{2}

5 0
3 years ago
If a coin has been altered so that the probability of heads coming up is 3/10, then the probability of tails coming up is also 3
soldier1979 [14.2K]
False. In a coin, there are 2 possible out comes i.e. a head and a tail. if p is the probability of one side coming up, then the probability of the other side comin up will be q = 1 - p.

Therefore, if the probability of head coming up is 3/10, the probability of tail coming up will be 1 - 3/10 = 7/10.
8 0
3 years ago
Ava flipped a coin 2 times. What are all the possible outcomes in the sample space? Let T represent the coin landing tails up an
daser333 [38]
The four outcomes are:

HH
HT
TH
TT

There are 4 outcomes that are possible. I've never seen anyone flip and edge, so that does not count.
5 0
3 years ago
Read 2 more answers
Evaluate (b+c) if a=2,b=-5 and c= -3
anastassius [24]
The answer would be -8
6 0
3 years ago
Read 2 more answers
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