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Iteru [2.4K]
3 years ago
6

A basketball player participates in a one-and-one free throw situation. The player has a 80% free throw average. What is the pro

bability the player gets 2 points?
Mathematics
2 answers:
jekas [21]3 years ago
6 0

Answer:

80%

Step-by-step explanation:

Svet_ta [14]3 years ago
4 0

Answer:

80%

Step-by-step explanation:

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The oil prices for 2014, rounded to the nearest dollar, were: 95, 101, 101, 102, 102, 106, 104, 97, 93, 84, 76, 59. what is the
kipiarov [429]
  • Interquartile Range (IQR) = Q_3-Q_1 , with Q_3 as the upper quartile and Q_1 as the lower quartile.

Firstly, rearrange the data so that it's in ascending order: \{59,76,84,93,95,97,101,101,102,102,104,106\}

Next, find the median:

\{59,76,84,93,95,\boxed{97,101,}101,102,102,104,106\}\\\\\frac{97+101}{2}\\\\\frac{198}{2}\\\\99

Now to find the lower quartile, find the "median" of the data set that's to the left of 99:

\{\overbrace{59,76,84,93,95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\{\overbrace{59,76,\boxed{84,93}95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\frac{84+93}{2}\\\\\frac{177}{2}\\\\88.5

Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:

\{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102

Now that we have the upper and lower quartile, subtract them:

102-88.5=13.5

<u>In short, the IQR of this data set is 13.5.</u>

8 0
3 years ago
a gas tank contains 5.3 gallons the capacity of the tank is 12.5 gallons how much will it cost to fill the tank at 2.60 per gall
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7.2 x 2.60

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A(b+c) = ab + ac

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Alexeev081 [22]

Answer:

33

Step-by-step explanation:

(3×3 )+(1/2 ×3×4) ×4 = 33

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