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matrenka [14]
2 years ago
6

Use your line plot and answer the question below

Mathematics
1 answer:
Wittaler [7]2 years ago
5 0

Answer:

hgvcffgebebddbb

Step-by-step explanation:

g

hhgffggggvgfgfgcccc

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Which of the following statements is no True?
bezimeni [28]

Answer:

correct answer is A. because the square root of 122 is 11.06 which means its between 11 and 12, not 10 and 11

Step-by-step explanation:

4 0
2 years ago
Plz help thank u for number 9 and 10
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It is 6 for number 9 glad I could help
7 0
3 years ago
Read 2 more answers
1. A bag contains four green gumballs, three blue gumballs, and two red gumballs, all the same size and shape.
Dahasolnce [82]

Answer:

Step-by-step explanation:

a. P (red) = <u>2/9</u>

b. P (blue) = 3/9 = <u>1/3</u>

c. P (green) = <u>4/9</u>

d. Green, because out of the three types of gumballs, the green gumballs are more in number, therefore, it is the one you are most likely to get.

7 0
2 years ago
Write the equation for the translation. y=cos x; translated 5 units right.
Rainbow [258]

Answer:

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8 0
3 years ago
Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

5 0
3 years ago
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