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2 years ago

12

Determine the number and type of complex solutions and possible real solutions for each of the following equations. 1) 2x2 + 5

x + 3 = 0
2) 4x3 – 12x + 9 = 0
3) 2x4 + x2 – x + 6 = 0

1 answer:

DanielleElmas [232]2 years ago

8 0

2) 4*3-12x+9 = 0 yup

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You toss a coin a randomly selecte a number from 0 to 9. What is the probability of getting tails and selecting a 9?

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First lets find the probability of landing tails. Because there is 1 way to get tails and 2 possible outcomes, the probability is 1/2.

Now, lets find the probability of selecting a 9. Since this includes 0, there are 10 possible outcomes and 1 way to get a 9, so the probability is 1/10.

To find them combined, multiply:

1/2 * 1/10
1/20
0.05

Hope this helps!

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3 years ago

Verify the identity:<br> tan(x+pi/2)=-cotx<br><br> please!!

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First we use sin(a+b)= sinacosb+sinbcosa
and cos(a+b)=cosa cosb -sinasinb

tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)
and sin(x+pi/2) = sinxcospi/2 + sinpi/2cosx =cosx,
<span>cos(x+pi/2) = cosxcospi/2- sinxsinpi/2= - sinx,
</span> because <span>cospi/2 =0, </span>and <span>sinpi/2=1

</span><span>=tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)= cosx / -sinx = -1/tanx = -cotx
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2 years ago

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

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Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, 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 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

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

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

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

So

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

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

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2 years ago

Write the relation as a set of ordered pairs.

ankoles [38]

Answer:

option B is the answer.

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2 years ago

Bonnie withdrew $1,200 from her savings account. She spent 1/5 of the money on travel arrangements. She spent 2/3 of the remaini

svetoff [14.1K]

Withdrew 1200
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leaving her with : 1200 - 240 = 960
spent 2/3 on hardware : 2/3(960) = 1920/3 = 640
leaving her with : 960 - 640 = 320
320 - lunch = 300.50
320 - 300.50 = lunch
19.50 = lunch <==

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2 years ago

Read 2 more answers