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3 years ago

URGENT PLZ HELP ME!!!

Mathematics

1 answer:

hichkok12 [17]3 years ago

4 0

Answer:

You would click at (0,-7)

Step-by-step explanation:

Definition of the minimum point:

"The minimum value of a function is the place where the graph has a vertex at its lowest point. In the real world, you can use the minimum value of a quadratic function to determine minimum cost or area."

Although this is not a quadratic, it still has a minimum point.

The minimum point here would be at it's lowest point

The minimum/lowest point is (0,-7)

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A triangle is graphed on a coordinate plane, with a vertex located at a point represented by (x,y).

Sophie [7]

The value of a is between 0 and 1.

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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

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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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3 years ago

Find the first three terms of the sequence below. Tn=n2+2n+9

klasskru [66]

$\\ \bull\tt\longrightarrow T_n=n^2+2n+9$

$\\ \bull\tt\longrightarrow T_1=1^2+2(1)+9=1+2+9=12$

$\\ \bull\tt\longrightarrow T_2=2^2+2(2)+9=4+4+9=17$

$\\ \bull\tt\longrightarrow T_3=3^2+2(3)+9=9+6+9=24$

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Answer:

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Step-by-step explanation:

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Answer:

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