Examine the equation. 4(x – 3) = 4x – 12 Which of the following is true? (Check all that apply.)

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

8 0

Answer:

It is a true statement.

Any input will result in an equivalent equation.

It is equivalent to an equation of the form a = a.

Step-by-step explanation:

on edgy

erma4kov [3.2K]3 years ago

6 0

Answer:

It is a true statement;

Any input will result in an equivalent equation;

It is equivalent to an equation of the form a = a.

Step-by-step explanation:

$4(x-3)=4x-12$

This equation has infinitely many solutions. Since any value for x on the left side and on the right side makes a valid equality. Equations like that have uncountable elements in their solution set, for x may assume infinite values.

In other words, that is to say:

It is a true statement;

Any input will result in an equivalent equation;

It is equivalent to an equation of the form a = a.

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A jar contains 12 red marbles numbered 1 to 12 and 6 blue marbles numbered 1 to 6. A marble is drawn at random from the jar. Fin

lora16 [44]

Answer:

$(a)P(Red)=\dfrac{2}{3}\\(b)P(Odd) =\dfrac{1}{2}\\(c)P(\text{Red or Odd Numbered})=\dfrac{5}{6}\\(d)P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

Step-by-step explanation:

Number of Red Marbles{1,2,3,4,5,6,7,8,9,10,11,12},n(R)=12

Number of Blue Marbles{1,2,3,4,5,6},n(B)=6

Total Number of Marbles, n(S)=6+12=18

(a)Probability that the Marble is Red

$P(R)=\dfrac{n(R)}{n(S)} =\dfrac{12}{18} =\dfrac{2}{3}$

(b)Probability that the marble is odd-numbered.

Number of Odd-Numbered Balls, n(O)=9

$P(Odd)=\dfrac{n(O)}{n(S)} =\dfrac{9}{18} =\dfrac{1}{2}$

(c)Probability that the marble is red or odd-numbered.

n(Red)=12

n(Odd Numbered marbles)=9

n(Red and Odd Numbered Marbles)=6

$P(\text{Red or Odd Numbered})=P(Red)+P(Odd\:Numbered)-P(\text{Red and Odd Numbered)}\\=\dfrac{12}{18} +\dfrac{9}{18}-\dfrac{6}{18} =\dfrac{15}{18}\\P(\text{Red or Odd Numbered})=\dfrac{5}{6}$ (d)Probability that the marble is blue or even-numbered.

n(Blue)=6

n(Even Numbered marbles)=9

n(Blue and Even Numbered Marbles)=3

$P(\text{Blue or Even Numbered})=P(Blue)+P(Even\:Numbered)-P(\text{Blue and Even Numbered)}\\=\dfrac{6}{18} +\dfrac{9}{18}-\dfrac{3}{18} =\dfrac{12}{18}\\P(\text{Blue or Even Numbered})=\dfrac{2}{3}$