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olga55 [171]
3 years ago
5

A bag contains 4 b l u e marbles, 3 w h i t e marbles, and 10 r e d marbles. If a marble is drawn from the bag, replaced, and an

other marble is drawn, what is the probability of drawing first a b l u e marble and then a r e d marble?
Mathematics
1 answer:
lawyer [7]3 years ago
4 0

Answer:

40/289

Step-by-step explanation:

We can see these two events are independent as Regina replaced marble after drawing one marble.

We can find possible outcomes by adding number of blue, red and yellow marbles.

possible outcomes= 4+3+10= 17

Let us find probability of getting a blue marble on first draw.

P( blue\ marble)= \frac{4}{4+3+10} = \frac{4}{17}

Now we will find probability of getting red marble.

P(red/marble)= \frac{10}{17}

We can find probability of getting a blue marble and then red marble by multiplying both probabilities.

P(\text{ Blue then red})=\frac{4}{17}* \frac{10}{17} =\frac{40}{289}

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Step-by-step explanation:

I said because Exponential describes a very rapid increase. ... Exponential is also a mathematical term, meaning "involving an exponent." When you raise a number to the tenth power, for example, that's an exponential increase in that number. the part involves exponent is the y part same number multiple to it self to get the next answer  while the x part one specific number to get the next number which is linear but since theyre asking about the relationship it’s neither because both were supposed to have the same relationship linea for x and y or exponential for x and y but they both are different.

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3 years ago
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3 years ago
4598(345+892•231)•5617(1232+987)/2=?
jeka57 [31]

Answer:

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Step-by-step explanation:

4598\left(345+892\cdot \:231\right)\cdot \:5617\cdot \frac{1232+987}{2}\\=4598\cdot \:5617\cdot \frac{1232+987}{2}\left(345+206052\right)\\=4598\cdot \:5617\cdot \frac{2219}{2}\left(345+206052\right)\\=4598\cdot \:206397\cdot \:5617\cdot \frac{2219}{2}\\=\frac{2219\cdot \:4598\cdot \:206397\cdot \:5617}{2}\\=\frac{1.18286\dots E16}{2}\\=5.91431e15

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