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solong [7]
3 years ago
10

Solve 15 ≥ -3x or 2/5x ≥ -2. A.) {x | x ≤ -5} B.) {x | x ≥ -5} C.) {all reals}

Mathematics
2 answers:
Lapatulllka [165]3 years ago
6 0
15 \geq -3x\quad|:(-3)\qquad\vee\qquad  \dfrac{2}{5}x \geq -2\quad|\cdot5\\\\\\
-5 \leq x\qquad\vee\qquad  2x \geq -10\quad|:2\\\\\\
-5 \leq x\qquad\vee\qquad  x \geq -5\\\\\\x\geq -5\qquad\implies\qquad\boxed{\{x | x \geq -5\}}

Answer B.
Hunter-Best [27]3 years ago
5 0

The correct answer is <u> C.) {all reals} </u>

Hope this helps!!! :D

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ipn [44]

Answer:

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Step-by-step explanation:

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Mashcka [7]

Answer: 5:4

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Simplify. 8c³d²/4cd²
loris [4]

Answer:

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6 0
2 years ago
A survey is made to determine the number of households having electric appliances in a certain city. It is found that 75% have r
Mashcka [7]

Answer:

The probability that a household has at least one of these appliances is 0.95

Step-by-step explanation:

Percentage of households having radios P(R) = 75% = 0.75

Percentage of households having electric irons P(I) = 65% = 0.65

Percentage of households having electric toasters P(T) = 55% = 0.55

Percentage of household having iron and radio P(I∩R) = 50% = 0.5

Percentage of household having radios and toasters P(R∩T) = 40% = 0.40

Percentage of household having iron and toasters P(I∩T) = 30% = 0.30

Percentage of household having all three P(I∩R∩T) = 20% = 0.20

Probability of households having at least one of the appliance can be calculated using the rule:

P(at least one of the three) = P(R) +P(I) + P(T) - P(I∩R) - P(R∩T) - P(I∩T) + P(I∩R∩T)

P(at least one of the three)=0.75 + 0.65 + 0.55 - 0.50 - 0.40 - 0.30 + 0.20  P(at least one of the three) = 0.95

The probability that a household has at least one of these appliances is 0.95

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3 years ago
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Viktor [21]

Answer:

9-8

Step-by-step explanation:

60 divided by 7 is 8.5

4 0
3 years ago
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