How did Bill spend longer

adell [148]

Hi there!

Since ST is a tangent to the circle, we can use the relationship: tangent squared = external secant segment x entire secant segment.

WORK:
(I will be using x in place of ST)
x^2 = 7(23 + 7)
x^2 = 7(30)
x^2 = 210
x = squareroot(210) or approximately 14.5 inches

Hope this helps!! :)

4 0

3 years ago

Plzz help e for brainliest if 2 people answer

Doss [256]

Answer:

B

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please provide an explanation for the last three questions.

zavuch27 [327]

There is no picture attached to the question??

5 0

3 years ago

Read 2 more answers

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

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

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