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Dmitrij [34]
3 years ago
14

Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. The probabi

lity that Helen makes the second free throw given that she made the first is 0.85. What is the probability that she makes both free throws?
Mathematics
1 answer:
CaHeK987 [17]3 years ago
3 0

Answer: 0.6375

Step-by-step explanation:

Let's assume that the event that she'll make the first shot is given as P(A) while making the second shot is P(B). Therefore, P(A/B) = 0.85

Therefore, the probability that she makes both free throws will be denoted as:

= 0.75 × 0.85

= 0.6375

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If you calculate separately for pink and red using above formula (909pink, 606red, which equals to the same 1515)

Hope that answers it for you..
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3 years ago
_____ less than a number is 51
Kryger [21]

Answer:

W number - blank is 51.

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I don't understand what the question is. What is the number unless you want something random then here:

Blank - 52

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3 0
3 years ago
Solve the inequality.<br> -1.5(4x+1) ≥ 45-25(x+1)
777dan777 [17]

Answer:

x ≥ 21.5/19

Step-by-step explanation:

-1.5(4x + 1) ≥ 45 - 25(x + 1)

-6x - 1.5 ≥ 45 - 25x - 25

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3 0
2 years ago
The list below shows the Highlight in inches of the player on a man college basketball team
boyakko [2]

We have the data set:

{79, 83, 82, 78, 79, 77, 80, 73, 82, 72}

This data set has 10 items.

To complete the data summary is helpful to sort the data:

{72, 73, 77, 78, 79, 79, 80, 82, 82, 83}

The minimum value is 72.

<em>The quartile divides the data set in 4 parts. </em>

<em>The first quartile is the value for which 25% of the data is below.</em>

<em>The second quartile is the value for which 50% of the data is below. The second quartile is equivalent to the median.</em>

<em>The third quartile is the value for which 75% of the data is below.</em>

In this data set, as its size is a even number, the quartile fall between two positions, so we will calculate the average of this numbers.

The first quartile will fall in the position 0.25*10=2.5. Then we can calculate the 1st quartile as the average between the 2nd and 3rd number of the data set.

This values are 73 (2nd position) and 77 (3rd position), so the average is:

Q_1=\frac{73+77}{2}=75

We can repeat this with the other quartiles:

\begin{gathered} Q_2=M=\frac{79+79}{2}=79 \\ Q_3=\frac{82+82}{2}=82 \end{gathered}

Then:

The first quartile is 75.

The median (second quartile) is 79.

The third quartile is 82.

The maximum value is 83.

7 0
1 year ago
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