4. Aliyah, Brenda and Candy share a sum of money in the ratio of 3:5:6. After
Candy gives $100 to Aliyah and $50 to Brenda, the ratio becomes 2 : 3:3.
(a) Suppose Aliyah has $3x at the start, express Candy's initial sum of money in
terms of x.
(b) Find the value of x.
(c) Hence, how much money does Brenda have in the end?

(a) Candy's initial sum as a terms of x is $6x

(b) x = $60

(c) $350

Step-by-step explanation:

The given parameters are;

The ratio in which Aliyah, Brenda and Candy share the sum of money = 3:5:6

The amount Candy later gives Aliyah = $100

The amount Candy later gives Brenda = $50

The new ratio of the sum of the shared money between Aliyah, Brenda and Candy = 2:3:3

(a) Whereby Aliyah has $3x at the start, we have;

Total sum of mony = Y

Amount of Aliyah's initial share = Y × 3/(3 + 5 + 6) = Y×3/14

Therefore, Y×3/14 = $3x

x = Y×3/14 ÷ 3 = Y/14

Amount of Candy's initial share = Y × 6/14

Therefore Candy's initial sum as a terms of x = $6x

(b) Given that Aliyah's and Candy's initial sum as a function of x are   $3x and $6x, therefore, in the ratio 3:5:6, Brenda's  initial sum as a function of x = $5x

Which gives;

Total amount of money = $14x

With

6x - 150, 3x + 100, and 5x + 50, the ratio =is 2:3:3

Therefore, we have;

14·x × 2/(2 + 3 + 3) = (6·x - 150)

14·x × 2/(8) = (6·x - 150)

14·x × 1/4 = (6·x - 150)

7·x/2 = (6·x - 150)

12·x - 300 = 7·x

12·x - 7·x = 300

5·x = 300

x = $60

(b) The final amount of money with Brenda = 5x + 50 = 5 × 60 + 50 = $350

The final amount of money with Brenda = $350.

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

happy to

pls mark as brainliest

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation