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omeli [17]
3 years ago
11

Use the Distance Formula to find the distance between D(2, 0) and E(8, 6).

Mathematics
1 answer:
Mazyrski [523]3 years ago
5 0

Answer:

The answer is

<h2>6 \sqrt{2}  \:  \:  \: or \:  \:  \: 8.50 \:  \:  \:  \: units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  }  \\

where

(x1 , y1) and (x2 , y2) are the points

From the question

The points are D(2, 0) and E(8, 6

The distance between them is

|DE|  =  \sqrt{ ({2 - 8})^{2}  +  ({0 - 6})^{2} }  \\  =  \sqrt{( { -  6})^{2}  + ( { - 6})^{2} }  \\  =  \sqrt{36 + 36}  \\  =  \sqrt{72}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  = 6 \sqrt{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\  = 8.4852813

We have the final answer as

6 \sqrt{2}  \:  \:  \: or \:  \:  \: 8.50 \:  \:  \:  \: units

Hope this helps you

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Answer:

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Step-by-step explanation:

i don't know what the given values for x are

but if you know them, here's an example

let's say x=4

y=(4)+2

y=6

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3 years ago
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Step-by-step explanation:

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3 years ago
Solve linear equations 1/2x+y=-6 and y=3/5x+5
ser-zykov [4K]
1/2x+y=-6   y=3/5x+5

substitute
1/2x+3/5x+5=-6

11/10x+5=-6

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3 years ago
An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do
vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

Then,

P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007

c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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3 years ago
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Approximately 3.9 gallons I’m sure of
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