a person invest in 9000 in a bank. the bank pays 5% interest compounded semi annually. to the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 14800 dollars

10.1 years.

Step-by-step explanation:

It is given that,

Principal = 9000

Rate of interest = 5%

No. of times interest compounded = 2 times in an year

Amount after certain time = 14800

The formula for amount:

where, P is principal, r is rate of interest, n is no. of times interest compounded in an year and t is time in years.

Substitute the given values in the above formula.

Taking log both sides.

Therefore, the required time is 10.1 years.

y =  9, z = 8

Step-by-step explanation:

Here, the given equations are,

150x + 80y + 12z = 2,466 (1)

380x + 95y + 48z = 5,419 (2)

210x + 130y + 60z = 3,960 (3)

4 × equation (1) - Equation (2),

220x + 225y = 4,445  (4)

5 × equation (1) - equation (3),

540x + 270y = 8,370

⇒ 270 y = 8370 - 540 x

⇒ y = 31 - 2 x

Substituting the value in equation (4),

220 x + 225 ( 31 - 2 x ) = 4445

220 x + 6975 - 450 x = 4445

- 230 x = - 2530

x = 11

By putting this value in equation (5),

we get, y = 31 - 22

y = 9

By putting the value of x and y in equation (1),

1650 + 720 + 12 z = 2466

2370 + 12 z = 2466

12 z = 96

z = 8