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Marina86
5 days ago
8

The probability that Jane will go to a ballgame (event A) on a Monday is 0.73, and the probability that Kate will go to a ballga

me (event B) the same day is 0.61. The probability that Kate and Jane both go to the ballgame on Monday is 0.52. From the given scenario, we can conclude that events A and B are . NextReset
Mathematics
2 answers:
AnnZ [3.8K]5 days ago
8 0

The answer for all you Edmentum/Plato users is

dependent events because P(A and B) does not equal P(A) * P(B)

Inessa [3.9K]5 days ago
4 0

Events A and B are termed independent when

Pr(A\cap B)=Pr(A)\cdot Pr(B),

if not, events A and B are classified as dependent.

The events A, B and A∩B are:

  • A - Jane plans to attend a ballgame on Monday;
  • B - Kate intends to go to a ballgame on Monday;
  • A∩B - Both Kate and Jane will be at the ballgame on Monday.

Pr(A)=0.73,\ Pr(B)=0.61,\ Pr(A\cap B)=0.52.\\ \\ Pr(A)\cdot Pr(B)=0.73\cdot 0.61=0.4453\neq 0.52=Pr(A\cap B).

Answer: events A and B are dependent


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Detailed explanation

This scenario involves motion with constant acceleration.

The relevant variables include the following.

\boxed{u \ or \ v_i = initial \ velocity}

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The equation we apply is:

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First, convert 0.50 mm to meters: \boxed{0.50 \ mm = 0.50 \times 10^{-3} \ m = 5.0 \times 10^{-4} \ m}

Solution steps:

Rearrange the formula to isolate acceleration (a).

\boxed{ \ v^2 = u^2 + 2ad \ }

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\boxed{ \ 2ad = v^2 - u^2 \ }

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Insert the given values into the rearranged equation.

\boxed{ \ a = \frac{(1.0)^2 - (0)^2}{2(5.0 \times 10^{-4})} \ }

\boxed{ \ a = \frac{1}{10 \times 10^{-4}} \ }

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\boxed{ \ a = 1 \times 10^{3}\ = 1,000 \ m/s^2 \ }

This yields the flea's acceleration as 1,000 m/s².

Additional resources

  1. Scientific notation: brainly.com/question/7263463
  2. Determining substance mass: brainly.com/question/4053884
  3. Conversion into cubic units: brainly.com/question/1446243

Keywords: flea, jumping, leg extension, takeoff velocity, initial speed, displacement, acceleration, constant acceleration, unit conversion

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