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torisob [31]
3 years ago
8

In a box, there are 4 red and 9 white balls. How many balls should be taken out of the box, so that there will be at least two r

ed balls among them?
Mathematics
1 answer:
creativ13 [48]3 years ago
7 0

Answer: 11 balls

Step-by-step explanation:

Because there are a total of 13 balls and 4 of them are red, if you are drawing out randomly you have to draw a minimum of 11 balls to guarantee that at least two red balls are among them

Brainliest is much appreciated!

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Olce Question<br> What type of notation uses these brackets {}?<br> A. Interval<br> B. Set
Sergio [31]

Answer:

Interval?

Step-by-step explanation:

5 0
3 years ago
For a function p(x), as |x|—&gt; infinity then p(x)—&gt; 3. Which of the following graphs could be p(x)?
VladimirAG [237]

Answer:

D

Step-by-step explanation:

A pe x the answer is d

7 0
3 years ago
A girls hair is 12 3/8 inches long if she had 2 1/2 inches cut off how long will her hair be ?
Darya [45]

Answer: 9 and 7eights

Step-by-step explanation: Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction. Which would equal, 9 and 7eights.

5 0
3 years ago
Solve for
faust18 [17]

Given equation : n(17+x)=34x−r.

We need to solve it for x.

Distributing n over (17+x) on left side, we get

17n + nx  = 34x - r.

Adding r on both sides, we get

17n+r + nx  = 34x - r+r.

17n + r + nx = 34x.

Subtracting nx from both sides, we get

17n + r + nx-nx = 34x-nx

17n + r = 34x -nx.

Factoring out gcf x on right side, we get

17x + r = x(34-n).

Dividing both sides by (34-n), we get

\frac{(17x + r)}{(34-n)} = \frac{x(34-n)}{(34-n)}

\frac{(17x + r)}{(34-n)} = x

<h3>Therefore, final answer is x=\frac{(17x + r)}{(34-n)}.</h3>
7 0
3 years ago
Read 2 more answers
A gumball machine has different flavors sour apple,grape,orange and cherry . There are six of each flavor. $.50 are put in the m
tamaranim1 [39]

Answer:

30/552

Step-by-step explanation:

In order to solve this problem you need to multiply the probability of getting grape for the first gumball with the probability of getting grape for the second gumball. Since there are 6 grape gumballs and a total of 24 gumballs (6*4). Then the probability of getting grape for the firs one is

\frac{6}{24}

Now there are only 5 grape gumballs available and one less in the total supply, therefore the probability of getting grape in the second try is

\frac{5}{23}

Finally we multiply them together to find the probability of getting two grapes in a row.

\frac{6*5}{24*23} = \frac{30}{552}

4 0
3 years ago
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