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ruslelena [56]
3 years ago
10

Is (0,3) a solution to the equation y = x + 3?​

Mathematics
2 answers:
kap26 [50]3 years ago
6 0
Yes because
X=0
Y=3
3=0+3
3=3
Anna11 [10]3 years ago
3 0
Yes,
y=x+3
if x=0,
y=0+3
y=3
therefore (0,3) is a solution
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Two numbers total 66 and have a difference of 16
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 \left \{ {{x+y=66} \atop {x-y=16}} \right. \\+-----\\Addition\ method\\\\2x=82\ \ \ |divide\ by\ 2\\\\
x=41\\y=66-x\\y=66-41=25\\\\Numbers\ are:\ 41\ and\ 25.
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3 years ago
Write an equation of the line that passes through the given points.<br> (-2,8) and (1, -1)
KIM [24]

/Answer:

Short Answer Type:

P1 - (-2,8)

P2 - (1,1)

Slope: (1 - 8) / (1 + 2)

=> -7/3

Therefore: 1 = -7/3 + c

=> 10/3 = c

<h2><u>Therefore equation: y = -7x/3 + 10/3</u> </h2>
6 0
3 years ago
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Nastasia [14]
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5 0
3 years ago
Read 2 more answers
What is the least common multiple of 156 and 321
jekas [21]

Hello  Hdolly81

<u><em>The LCM of 156 and 321 is 16692</em></u>

Backup:

156: 2,2,3,13,

321: 3,107

2 * 2 *  3 * 13 * 107 = 16692

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5 0
3 years ago
The random variable x has the following probability distribution: x f(x) 0 .25 1 .20 2 .15 3 .30 4 .10 a. Is this probability di
Reptile [31]

Answer and Explanation:

Given : The random variable x has the following probability distribution.

To find :

a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.

b. Calculate the expected value of x.

c. Calculate the variance of x.

d. Calculate the standard deviation of x.

Solution :

First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.25           0               0                0

1     0.20        0.20             1              0.20

2    0.15          0.3               4             0.6

3    0.30         0.9               9             2.7

4    0.10          0.4               16             1.6

   ∑P(x)=1     ∑xP(x)=1.8               ∑x²P(x)=5.1

a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.25+0.20+0.15+0.30+0.10

\sum P(X)=1

Yes it is a probability distribution.

b) The expected value of x is defined as

E(x)=\sum xP(x)=1.8

c) The variance of x is defined as

V=\sum x^2P(x)-(\sum xP(x))^2\\V=5.1-(1.8)^2\\V=5.1-3.24\\V=1.86

d) The standard deviation of x is  defined as

\sigma=\sqrt{V}

\sigma=\sqrt{1.86}

\sigma=1.136

5 0
3 years ago
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