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3 years ago

Which statements are true? Check all that apply.

Mathematics

2 answers:

Basile [38]3 years ago

7 0

Answer:

True, 1, 3, 5

Step-by-step explanation:

1. Angle L is sircumsribed angle, thenMJ⊥JL and MK⊥KL, so the measure of the angle L is

360°-131°-90°-90°=49° (the sum of all interior angles in the quadrilateral is 360°)

True

2. False, because m∠L=49°.

3. If the measure of arc HJ is 98°, then the measure of the central angle HMJ is 98°.

True

4. If the measure of arc HJ is 98°, then the measure of the central angle HMJ is 98° and the meausre of inscribed angle HKJ subtended on the arc HJ is half of the central angle measure and is equal to 49°.

False

5. If the measure of arc HJ is 98°, then the measure of the arc HK is

360°-131°-98°=131°

True

Maru [420]3 years ago

3 0

The circumscribed angle L has a measure of 49°.

If the measure of arc HJ is 98°, the measure of angle HMJ is 98°.

If the measure of arc HJ is 98°, the measure of arc HK is 131°.

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Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 18 means that $N = 18$

9 are carrying nuclear weapons, which means that $k = 9$

9 are destroyed, which means that $n = 9$

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021$

$P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687$

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

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Answer:

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