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kirill [66]
3 years ago
13

What is the answer for this problem for Solving Two Step Equations?

Mathematics
1 answer:
avanturin [10]3 years ago
4 0
X=-12

or x=12 is 4x is positive
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I need help on these two questions please
Simora [160]

Answer:

<h2>For #1=     D = 24.25</h2><h2>     #2=       D= 21</h2>

Step-by-step explanation:

hope this helps. pls mrk me branliest :)

4 0
3 years ago
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Perform the indicated operation.<br> (w^3 +64)/(4+ w)
timurjin [86]

Answer:

(w^2 - 4w + 16)

Step-by-step explanation:

Note that w^3 +64 is the sum of two perfect cubes, which are (w)^3 and (4)^3.  The corresponding factors are (w + 4)(w^2 - 4w + 16).

Therefore,

(w^3 +64)/(4+ w) reduces as follows:

   (w^3 +64)/(4+ w)      (4 + w)(w^2 - 4w + 16)

--------------------------- = --------------------------------- =  (w^2 - 4w + 16)

             4 + w                     4 + w

6 0
3 years ago
What property is 5x6x7=5x7x6
stich3 [128]

Commutative Property

5 0
3 years ago
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Help I don't know what I did incorrectly, can you tell me then give me the answer
mafiozo [28]

Answer:

the slope is negative since its going downward

Step-by-step explanation:

3 0
2 years ago
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Suppose that 10% of all homeowners in an earthquake-prone area of California are insured against earthquake damage. Four homeown
guapka [62]

Remainder of question:

Find the probability distribution of x

Answer:

The random variable x is defined as: X = {0, 1, 2, 3, 4}

The probability distribution of X:

P(X = 0) = 0.656

P(X = 1) = 0.2916

P(X= 2) = 0.0486

P(X=3) = 0.0036

P(X = 4) = 0.0001

Step-by-step explanation:

Sample size, n = 4

Random variable, X = {0, 1, 2, 3, 4}

10% (0.1) of the homeowners are insured against earthquake, p = 0.1

Proportion of homeowners who are not insured against earthquake, q = 1 - 0.1

q = 0.9

Probability distribution of x,

P(X = r) = ^nC_r *p^r q^{n-r} \\\\P(X= 0) =(^4C_0 *p^1 q^4 )\\P(X=0) = (^4C_0 *0.1^0 0.9^4 ) = 0.656\\P(X= 1)= (^4C_1 *p^1 q^3 )\\P(X=1) = (^4C_1 *0.1^1 0.9^3 ) = 0.2916\\P(X= 2)=( ^4C_2 *p^2 q^2) \\P(X=2) = (^4C_2 *0.1^2 0.9^2 ) = 0.0486\\P(X= 3) = (^4C_3 *p^3 q^3) \\ P(X=3) = (^4C_3 *0.1^3 0.9^1 ) = 0.0036\\P(X= 4) =  (^4C_4 *p^4 q^0 )\\ P(X=4) =(^4C_4 *0.1^4 0.9^0 ) = 0.0001

5 0
3 years ago
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