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Brilliant_brown [7]
3 years ago
11

An athlete plains to run 3 miles. Each lap around th school yard is 3/7 miles how many laps will the athlete run

Mathematics
1 answer:
kati45 [8]3 years ago
3 0

Answer: 7 laps

Step-by-step explanation:

The athlete planned to run 3 miles

1 lap = 3/7 miles

x laps = 3 miles

by proportion :

x × 3/7miles = 3 miles

x = 3 ÷ 3/7

x = 3 x 7/3

x = 7

Therefore: the athlete will run 7 laps

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Solve for f in the proportion.<br> 28/44 = F/99 F = ?
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Answer:

F = 63

Step-by-step explanation:

28/44 = F/99

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3 years ago
What is the solution to this system of linear equations? Y-x=6 y+x=-10
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8 0
3 years ago
A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o
balu736 [363]

Answer:

a

P(X \ge 1) = 0.509

b

P(X  \ge 1) = 0.6807

Step-by-step explanation:

From the question we are told that

   The number of students in the class is  N  =  20  (This is the population )

   The number of student that will cheat is  k =  3

   The number of students that he is focused on is  n  =  4

Generally the probability distribution that defines this question is the  Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally  probability mass function is mathematically represented as

      P(X = x) =  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}

Here C stands for combination , hence we will be making use of the combination functionality in our calculators  

Generally the that  he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

      P(X \ge 1) =  1 - P(X \le 0)

Here  

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{20 - 3} C_{4- 0}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{17} C_{4}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ 1 *  2380}{ 4845}

    P(X \le 0) =  0.491

Hence

    P(X \ge 1) =  1 - 0.491

     P(X \ge 1) = 0.509

Generally the that  he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

    P(X \ge 1) =  1 - P(X \le 0)

   P(X  \ge 1) =1- [  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]

Here n =  6

So

    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{1  *  12376}{38760}]

    P(X  \ge 1) =1- 0.3193

    P(X  \ge 1) = 0.6807

   

5 0
3 years ago
(x-3)(x+6)<br>multiply and simplify your answer ​
enot [183]

Answer:

x^{2} + 9x +18

Step-by-step explanation:

multiply the parentheses together (x × x)=x² and x×6=6x then do 3×x which will equal 3x and then multiply 3 and 6 which is 18

now combine like terms (x²+6x+3x+18)

6+3=9 so now you will have x²+9x+18

5 0
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