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Bas_tet [7]
3 years ago
13

Which technique is most appropriate to use to solve each equation? Drag the name of the technique into the box to match each equ

ation. (x+3)(x+2)=0 ; x² + 6 = 31
1.) zero product property

2.) square root property

3.) quadratic formula
Mathematics
2 answers:
attashe74 [19]3 years ago
8 0

Answer:

(x+3)(x+2)=0-----zero product property

x² + 6 = 31 =-----quadratic formula

Step-by-step explanation:

THe zero product property you take an equation that is product of the multiplication of two binomials and by factorizing it you take both factors and equalize it to 0 in order to find the value of both values for x.

The quadratic formula is a formula used to solve quadratic equations that can´t be determined by the square root property or the zero product. You just use the numerical terms of the the formula:

Ax^{2} +Bx+C=0

And the formula is:

x=x=\frac{-B+\sqrt{B^{2}-4AC } }{2A}

And you just put the values of the quadratic function into the formula.

faltersainse [42]3 years ago
5 0
(x + 3) (x + 2) = 0
 To solve it, the most appropriate technique is:
 1.) zero product property
 The solutions are:
 (x + 3) = 0
 x = -3
 (x + 2) = 0
 x = -2
 
 x² + 6 = 31
 
To solve it, the most appropriate technique is:
 2.) square root property
 x² = 31-6
 x² = 25
 x = +/- root (25)
 x = +/- 5
 The solutions are:
 x = 5
 x = -5
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Answer:

x = -2 and x = -4

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2 years ago
A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with t
zaharov [31]

Answer:

a. 0.2

b. 0.42

c. 0.7

d. the solution is in the explanation

e. x and y are not independent

Step-by-step explanation:

a. from the joint probability mass function table,

p(x=1) and p(Y= 1)

= p(1,1) = 0.2

b. prob(0,0)+prob(0,1)+prob(1,0)+prob(1,1)

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= 0.42

P(X ≤ 1 and Y ≤ 1) = 0.42

c. prob {X ≠ 0 and Y ≠ 0}

= prob(1,1) + prob(1,2) + prob(2,1) + prob(2,2)

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d. we have to calculate the marginal pmf of x and y here.

<u>we have the x values as 0,1,</u>2

prob(x=0) = 0.1 + 0.04 + 0.02

= 0.16

prob(x=1) = 0.08 + 0.2 + 0.06

= 0.34

prob(x=2) = 0.06+0.14+0.3

= 0.50

<u>we have y values as 0,1,2</u>

prob(y=0) = .1+.08+.06

= 0.24

prob(y=1) = .04+.2+.14

= 0.38

prob(y = 2) = 0.02+0.06+0.3

= 0.38

P(X ≤ 1) = prob(x=0)+prob(x=1)

= 0.34+0.16

= 0.50

e. from the joint table we have this,

prob(1,1) = 0.2

prob(x=1) = 0.34

prob(y=1) = 0.38

then prob(x=1)*prob(y=1)

= 0.34*0.38

= 0.1292

therefore prob(1,1) is not equal to prob(x=1)*prob(y=1)

0.2≠0.1292

x and y are not independent

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Answer:

Follows are the solution to this question:

Step-by-step explanation:

Given:

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Step-by-step explanation:

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