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Alecsey [184]
2 years ago
8

Which image is a reflection over the line y = x of triangle 1?

Mathematics
2 answers:
balu736 [363]2 years ago
8 0

Answer:

a

Step-by-step explanation:

rjkz [21]2 years ago
5 0

Answer:

A edge 2021

Step-by-step explanation:

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What is the solution to the system of equations below? x 3 y = 15 and 4 x 2 y = 30 (6, 3) (3, 6) (7, –6) (–6, 7).
swat32

Answer: (6 , 3)

Step-by-step explanation: Given

x + 3y =15 ------- equation 1

4x + 2y = 30 --------- equation 2

x = 15 - 3y

Put value of x in equation 2

4 (15 - 3y) + 2y = 30

60 - 12y + 2y = 30

-10y = -30

divide by '-10' on both sides

y = 3

now, put value of y in equation 1

x + 3(3) = 15

x + 9 = 15

x = 15 - 9

x =  6

So , (x , y) = (6 , 3)

7 0
2 years ago
If g(x) = 4x + 1 and g(x) = -11, find x.
podryga [215]

Answer:

x = - 3

Step-by-step explanation:

g(x) = 4x + 1...... (1)

g(x) = -11.... (2)

From equations (1) & (2)

-11 = 4x + 1

-11 - 1 = 4x

-12 = 4x

-12/4 = x

-3 = x

x = - 3

3 0
2 years ago
A number that multiplies a variable in a term is a _-
Leto [7]
It is called a variable
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3 years ago
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The inequality is < because the equal sign is an equality.
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2 years ago
Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste
lidiya [134]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

For this case we have the following distribution given:

X          3      4       5        6

P(X)   0.07  0.4  0.25  0.28

We can calculate the mean with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74

In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36

And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

3 0
3 years ago
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