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3 years ago

Inflation causes things to cost more, and for our money to buy less. Suppose inflation decreases the value of money by 1.1% each

Mathematics

1 answer:

Andreas93 [3]3 years ago

8 0

Answer: $171.31

Step-by-step explanation:

A = P(1 - r)ⁿ

Given: P = 200, r = 1.1% (.011), n = 14

A = 200(1 - .011)¹⁴

= 200(.989)¹⁴

= 171.308

Since we are talking about money, round to two decimal places.

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What does a tape diagram look like

Sergeu [11.5K]

Here is an example

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3 years ago

Kelsey went bowling. It cost her a one-time fee of $3.50 to rent shoes, and $6.50 per game she played. She does not want to spen

Alla [95]

$6.50g+$3.50=$23.00

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3 years ago

What is the solution to this equation: 7m to the 2 power + 6m - 1 = ?

Mumz [18]

Answer:

7m² + 6m - 1

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Combining Like Terms
Standard Form: ax² + bx + c = 0

Step-by-step explanation:

Step 1: Define expression

7m² + 6m - 1

Step 2: Simplify

The expression cannot be simplified further.

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3 years ago

The table shows the temperatures in degrees Celsius, y, inside a shed at x hours past noon. The maximum and minimum temperatures

kolbaska11 [484]

Answer:

y = 3 cos (pi/12x - pi/4) + 17

Step-by-step explanation:

First find the average temperature which is 20 + 14/2 = 17

Find the amplitude

20 - 17 = 3

This means that that at noon, the shift is 3hrs past

the best equation that represents this scenario is $y = 3 cos (\frac{\pi}{12}x - \frac{\pi}{4}) + 17$

5 0

3 years ago

A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time

user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

$a(t) = 24t + 2$

Velocity:

$v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K$

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

$v(t) = 12t^{2} + 2t + 13$

Position:

$s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K$

Since s(0) = 3

$s(t) = 4t^{3} + t^{2} + 13t + 3$

What is its position at time t=5?

This is s(5).

$s(t) = 4t^{3} + t^{2} + 13t + 3$

$s(5) = 4*5^{3} + 5^{2} + 13*5 + 3$

$s(5) = 593$

It's position at time t = 5 is 593.

3 0

3 years ago