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vesna_86 [32]
3 years ago
12

Given that there were 4 Heads in the first 7 tosses, find the probability that the 2nd Heads occurred at the 4th toss. Give a nu

merical answer.
Mathematics
1 answer:
tamaranim1 [39]3 years ago
6 0

Answer:

The probability that the 2nd Heads occurred at the 4th toss is 5.99%.

Step-by-step explanation:

Given that there were 4 Heads in the first 7 tosses, to find the probability that the 2nd Heads occurred at the 4th toss the following calculation must be performed:

7 - 4 = 3

4/7 x 3/7 x 3/7 x 4/7 = X

0.571 x 0.428 x 0.428 x 0.571 = X

0.0599 = X

Therefore, the probability that the 2nd Heads occurred at the 4th toss is 5.99%.

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NemiM [27]

Answer:

y=2x*6

Step-by-step explanation:

5 0
3 years ago
what is the volume of the prism given that each cube is defined by the fractional edge length of 1/3 inch
Diano4ka-milaya [45]
The equation for volume is length times width times height. So to solve this multiple 1/3x1/3x1/3 because 1/3 is the length, width , and height of the cube(cubes have equal sides). The answer should be 1/27 
3 0
3 years ago
**22 POINTS** Please solve this fraction as a difference
Ghella [55]

Answer:

\dfrac{x}{4}-\dfrac{7}{12}

Step-by-step explanation:

The fraction is the equivalent of ...

\dfrac{1}{12}{(3x-7)

and the distributive property applies.

=\dfrac{1}{12}(3x)-\dfrac{1}{12}(7)\\\\=\dfrac{3\cdot x}{3\cdot 4}-\dfrac{7}{12}\\\\=\dfrac{x}{4}-\dfrac{7}{12}

3 0
3 years ago
heather has divided $6300 between two invesments, one paying 9%, the other paying 4%. If the return on her investment is $372, h
Ganezh [65]
Assuming that the two investments are  X & Y
X + Y = 6300
X = 6300 - Y                                       (1)

9/100X + 4/100 Y = 372                       (2)
replacing X from (1) into (2)
9/100(6300-Y) + 4/100 Y = 372
567 - 9/100Y +4/100Y = 372
(-9+4)/100Y = 372 - 567
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I hope this is helpful 



6 0
3 years ago
A survey of 46 college athletes found that 24 played volleyball, while 22 played basketball. a) If we pick one athlete survey pa
PSYCHO15rus [73]

Answer:

A) \dfrac{11}{23}

B) \dfrac{88}{345}

Step-by-step explanation:

A survey of 46 college athletes found that

  • 24 played volleyball,
  • 22 played basketball.

A) If we pick one athlete survey participant at random,  the probability they play basketball is

P_1=\dfrac{22}{46}=\dfrac{11}{23}

B) If we pick 2 athletes at random (without replacement),

  • the probability we get one volleyball player is \dfrac{24}{46}=\dfrac{12}{23};
  • the probability we get another basketball player is \dfrac{22}{45} (only 45 athletes left).

Thus, the probability we get one volleyball player and one basketball player is

P_2=\dfrac{12}{23}\cdot \dfrac{22}{45}=\dfrac{88}{345}

5 0
3 years ago
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