In a fish tank, 65% is female fishes. the rest are male. female fishes are also 15 less than twice of male fishes. how many female fishes and male fishes in the tank ? (set up the rational equation and solve for x)

Step-by-step explanation:

Let x be the number of male fish. Then 2x-15 is the number of female fish, and the ratio of female to male fish is ...

  0.65/0.35 = (2x -15)/x . . . . . rational equation

  13/7 = 2 - 15/x . . . . . . . . . . . . simplified

  15/x = 2 -13/7 = 1/7 . . . . . . . . add 15/x - 13/7

  x = 7·15 = 105 . . . . . . . . . . . . multiply by 7x

  2x-15 = 210 -15 = 195 . . . . . . use the expression for the number of females

There are 195 female fish and 105 male fish in the tank.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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