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3 years ago

Help me please help please

Mathematics

1 answer:

Vadim26 [7]3 years ago

3 0

Answer:

x = 2

Step-by-step explanation:

there is no y intercept

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Okay have posted the next question.

Verdich [7]

Answer:

1. only 1 unique triangle can be made

2. only 1 unique triangle can be made

3. multiple triangles are possible

4. True

5. True

6. False

7. only 1 unique triangle can be made

8. only 1 unique triangle is possible

9. multiple triangles are possible

10. no triangle is possible

Step-by-step explanation:

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3 years ago

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Name at least two independent variables that could result in a change in the price of a basket of apples.

Novay_Z [31]

Answer:

In an experimental study looking at classical music exposure and reading ability in children, the researcher divided the children into two groups (Groups A and B

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2 years ago

PLZ HELP ASAP DUE RN

matrenka [14]

Answer:

The first and third options are correct so choose both

Step-by-step explanation:

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2 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:

$P(X \ge 1) = 0.509$

$P(X \ge 1) = 0.6807$

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

$P(X \ge 1) = 1 - P(X \le 0)$

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