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mafiozo [28]
3 years ago
11

What percent of 965 is 1932

Mathematics
2 answers:
ankoles [38]3 years ago
4 0

Answer: It is 202.1 lol

Lena [83]3 years ago
4 0
The correct answer is 200.21%
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The voltage in a circuit is the product of two factors, the resistance and the current. If the voltage is 6ir + 15i + 8r+20, fin
zhuklara [117]

Answer:

  • resistance: (2r +5)
  • current: (3i +4)

Step-by-step explanation:

The factors of the given expression are ...

  6ir +15i +8r +20 = (3i +4)(2r +5)

Which factor is current and which is resistance is not clear. Usually, resistance is referred to using the variable r, so we suppose the expressions are supposed to be ...

  resistance: (2r +5)

  current: (3i +4)

6 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
If f(x)=2x, then f^-1(x)=
-BARSIC- [3]

f(x)=2x\to y=2x\\\\\text{change x and y}\\\\x=2y\\\\\text{solve for y}\\\\2y=x\qquad|:2\\\\y=\dfrac{x}{2}\\\\Answer:\ f^{-1}(x)=\dfrac{x}{2}

8 0
3 years ago
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