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2 years ago

Which statement is NOT a step used when constructing an inscribed hexagon?

Mathematics

1 answer:

den301095 [7]2 years ago

3 0

Answer:

Swing two arcs above and below the radius.

Step-by-step explanation:

We want to determine which statement is NOT a step used when constructing an inscribed hexagon?

1. Swing two arcs above and below the radius. False

2. Draw a radius of the circle using a straightedge. True

3. Place the compass on the point where the circle and radius intersect. True

4. Swing an arc the length of the radius from the point on the circle. True

The first choice is the correct answer.

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Round to underline digit 3 in 32, 336

Charra [1.4K]

I dont know if you know this, but there are 3 different 3's. So how can we figure out the question if we don't know which one is underlined?

5 0

3 years ago

Please answer this question

zimovet [89]

Answer:

The answer is 030

Step-by-step explanation:

I am unable to transmit my working but don't forget to write the bearing as 030 not 30

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2 years ago

What number added to 66 is 167

denis-greek [22]

Answer:

101

Step-by-step explanation:

To figure this out, you would do 167 - 66 = 101

101 + 66 = 167

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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

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

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