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SIZIF [17.4K]
3 years ago
11

What do all rectangles have in common? Choose ALL the correct answers.

Mathematics
2 answers:
Yakvenalex [24]3 years ago
5 0

Answer:

4 right angles

Step-by-step explanation:

All rectangles have <u>4 right angles</u> in common.

natali 33 [55]3 years ago
4 0

Answer:

4 right angles!!!

Step-by-step explanation:

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Step-by-step explanation:

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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
Can someone help me please i need help
DerKrebs [107]

Answer:

71

Step-by-step explanation:

5 0
3 years ago
Use parentheses and multiplication properties then find the product 9x1x5=
Delicious77 [7]

im guessing this is what you're asking for

9(1(5))

the answer is 45

6 0
3 years ago
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