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vazorg [7]
3 years ago
7

The coordinates of the verticies of triangle ABC are A(2,5), B(6,-1), and C(-4,-2). Fint the perimeter of triangle ABC, to the n

earest tenth
Mathematics
1 answer:
deff fn [24]3 years ago
8 0

Answer:

24.4 Units

Step-by-step explanation:

For this, we will use the distance formula to find the distance between each vertex

\sqrt({x_{1}-x_{2} })^{2} +(y_{1} -y_{2})^{2}

So we do this for each side of the triangle to get AB = \sqrt{52}, BC = \sqrt{109}, and AC = \sqrt{45}

We add these all up to get 24.4 (rounded)

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The answer is 41.6 or 42

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3 years ago
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What is the value of x ?
statuscvo [17]
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Hope this helps!
8 0
3 years ago
The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.9 mg and standard deviation 0.1
Ad libitum [116K]

Answer:

The answer to the question is;

The probability that the resulting sample mean of nicotine content will be less than 0.89 is 0.1587 or 15.87 %.

Step-by-step explanation:

The mean of the distribution = 0.9 mg

The standard deviation of the sample = 0.1 mg

The size of the sample = 100

The mean of he sample = 0.89

The z score for sample mean is given by

Z =\frac{X-\mu}{\sigma/ \sqrt{n} } where

X = Mean of the sample

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σ = Standard deviation of the population

Therefore Z = \frac{0.89-0.90}{0.1/\sqrt{100} } = -1

From the standard probabilities table we have the probability for  a z value of -1.0 = 0.1587

Therefore the probability that the resulting sample mean will be less than 0.89 = 0.1587 That is the probability that the mean is will be less than 0.89 is 15.87 % probability.

             

7 0
3 years ago
the radius of a gold atom is 1.35 angstroms. how many gold atoms would it take to line up a span of 8.5 mm?
serg [7]
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diameter = 2 x 1.35 x10^ -10 m ( 1000 mm / 1 m ) = 2.7 x 10^-10 mm / atom

Number of gold atoms = 8.5 mm / </span>2.7 x 10^-10 mm / atom = 31481481481.481 atoms
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3 years ago
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laila [671]
3x+2=x+8
Slandered Form
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Factorizations
2(x-3)=0
Solutions
x=6/2=3  (6 divided by 2 = 3)
    
6 0
3 years ago
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