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iVinArrow [24]
3 years ago
12

Juan made 13 out of 20 free throws. If Botina shoots 25 free thows, what's the minimum number she has to make in order to have a

better free-throw percentage than Juan?​
Mathematics
2 answers:
Dahasolnce [82]3 years ago
7 0

Answer:

the answer is 17

Step-by-step explanation:

juans free throw % is 65 so botina has to make 17 out of 25 to get a better percent which is 68

Lorico [155]3 years ago
7 0

<u>ANSWER:  </u>

Botina has to make 17 successful attempts out of 25 attempts.

<u>SOLUTION: </u>

Given, Juan made 13 out of 20 free throws.

Botina shoots 25 free throws,  

We need find what's the minimum number she has to make in order to have a better free-throw percentage than Juan?

Let, the minimum number of successful throws be x.

Now, percentage of Botina should be greater than percentage of jaun

Percentage of Botina > percentage of Jaun

\begin{array}{l}{\frac{\text {number of successful attempts by botina}}{\text {total number of attempts by botina}} \times} {100>\frac{\text {number of successful atempts by jaun}}{\text {total number of attempts by jaun}} \times 100}\end{array}

\begin{array}{l}{\frac{x}{25} \times 100>\frac{13}{20} \times 100} \\\\ {\frac{x}{25}>\frac{13}{20}} \\\\ {x>25 \times \frac{13}{20}} \\\\ {x>5 \times \frac{13}{4}} \\\\ {x>\frac{65}{4}} \\\\ {x>16.25}\end{array}

As the number of attempts cannot be in decimal values, we have to consider the integer above 16.25 i.e. 17

Hence, botina has to make 17 successful attempts.

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At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more
I am Lyosha [343]

Answer:

The probability is  P( X \le 4 ) = 0.0054

Step-by-step explanation:

From the question we are told that

   The percentage that are on time is  p =  0.76

   The  sample size is n =  11

   

Generally the percentage that are not on time is

     q =  1- p

     q =  1-  0.76

     q = 0.24

The  probability that no more than 4 of them were on time is mathematically represented as

        P( X \le 4 ) =  P(1 ) +  P(2) + P(3) +  P(4)

=>     P( X \le 4 ) =  \left n } \atop {}} \right.C_1 p^{1}  q^{n- 1} +   \left n } \atop {}} \right.C_2p^{2}  q^{n- 2} +  \left n } \atop {}} \right.C_3 p^{3}  q^{n- 3}  +  \left n } \atop {}} \right.C_4 p^{4}  q^{n- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{11- 1} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{11- 2} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{11- 3}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{11- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{10} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{9} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{8}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{7}

= \frac{11! }{ 10! 1!}  (0.76)^{1}  (0.24)^{10} +   \frac{11!}{9! 2!}  (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!}  (0.76)^{3}  (0.24)^{8}  + \frac{11!}{7!4!}  (0.76)^{4}  (0.24)^{7}

P( X \le 4 ) = 0.0054

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erma4kov [3.2K]

Answer:

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Step-by-step explanation:

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