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3 years ago

Someone please help me!!

Mathematics

1 answer:

eduard3 years ago

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The improper fraction 9/8 equals what mixed number

Lana71 [14]

The answers is 1 1/8

4 0

3 years ago

Please help me quickly!

Degger [83]

Answer:

√50

Step-by-step explanation:

Use Pythagorean Theorem: a^2+b^2=c^2

5^2+5^2=c^2

25+25=c^2

50=c^2

c=√50

3 0

2 years ago

Read 2 more answers

Find the area of this polygon

Levart [38]

<h3>Answer: 12 square units</h3>

Explanation:

Rectangle ABDE is 4 units across the horizontal, and 2 units tall.

The area of this rectangle is length*width = 4*2 = 8 square units.

Triangle BCD has a base of 4 and height 2. The area of which is base*height/2 = 4*2/2 = 4 square units.

The total area is

rectangle + triangle = 8 + 4 = 12 square units

8 0

2 years ago

!!!! PLEASE HELP !!!!!!

Keith_Richards [23]

Answer:

i got -8 but the closest is -7 so u can tried that

Step-by-step explanation:

5 0

3 years ago

Naval intelligence reports that 99 enemy vessels in a fleet of 1818 are carrying nuclear weapons. If 99 vessels are randomly tar

oee [108]

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.

4 0

3 years ago