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-Dominant- [34]
2 years ago
8

What is the nature of triangle formed by the lines x+y=0 , 3x+y=0 , x+3y-4=0

Mathematics
1 answer:
Marrrta [24]2 years ago
4 0

Answer:

  isosceles obtuse triangle

Step-by-step explanation:

The attached graph shows a plot of the three lines.

The triangle is an isosceles obtuse triangle in the 2nd quadrant.

_____

It is convenient to let a graphing calculator plot these lines. If you're going to plot them by hand, it usually works best to choose values of the variables for the variable that has the largest coefficient (3 in two of the equations). That way, you're not trying to plot coordinates that contain fractions.

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Answer:

I also got 27.44 mg.

Step-by-step explanation:

Use the function given to you:

A(t)=le^rt

l: Initial Amount (50)

r: Rate (-.1) (Negative because the medicine is exiting the bloodstream at 10% per hour

t: 6 (the specific time value you are evaluating the function at.

A(6)=50e^-.1(6)=27.44

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3 years ago
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Answer:

3 dollars for each can of tuna

Step-by-step explanation:

9-3=6

6-3=3

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-3,2   -2,4  1,-1

Step-by-step explanation:

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In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability th
MrRa [10]

Answer:

a) 0.984

b) 20 people

Step-by-step explanation:

a)

If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.

Now, we want to know how many samples there are with this property.

Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is

\frac{20!}{18!2!}=190

In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely

(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

where 1=carries the gene, 0=does not carry the gene.

So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.

b)

Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.

4 0
3 years ago
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