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alex41 [277]
3 years ago
13

Suppose you deposit $2,000 in a savings account that pays interest at an annual rate of 6%. If no money is added or withdrawn fr

om the account, answer the following questions.
a. How much will be in the account after 4 years?
b. How much will be in the account after 17 years?
c. How many years will it take for the account to contain $2,500?
d. How many years will it take for the account to contain $3,000?
Mathematics
1 answer:
aleksandrvk [35]3 years ago
7 0

Answer:

  • a. $2480
  • b. $4040
  • c. 4 years and 2 months
  • d. 8 years and 4 months

Step-by-step explanation:

a. Account after 4 years:

  • 2000 + 4*2000*0.06 = $2480

b. Account after 17 years:

  • 2000 + 17*2000*0.06 = $4040

c. Time when account = $2500:

  • 2000 + x*2000*0.06 = 2500
  • 120x = 500
  • x = 500/120
  • x = 4 20/120 = 4 2/12 years = 4 years and 2 months

d. Time when account = $3000

  • 2000 + x*2000*0.06 = 3000
  • 120x = 1000
  • x = 1000/120
  • x = 8 40/120 = 8 4/12 years = 8 years and 4 months
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Can someone please help me with these 3, 7th grade math homework word problems. I will give brainliest and obviously a thanks
NikAS [45]

Answer:

Mia can spend no more than $55 at the mall. She buys a jacket for $23. Necklaces are on sale for $8. Find the amount of necklaces Mia can buy. Write and solve an inequality.

First, make the equation. She can't waste no more than(≥ or =) $55. And she already used $23.

55= 23+8(n)

n= number of necklaces she can buy.

Now, subtract 23 to both sides.

55= 23+8(n)

-23  -23

32= 8(n)

Then, divide by 8 to both sides to find how many necklaces she can buy.

<u>32</u>= <u>8(n)</u>

8      8

4=n So, she can buy 4 necklaces.

A bag of marbles contains 3 red, 5 blue and 2 yellow marbles. What is the probability of choosing a blue marble and then choosing a red marble without replacement. Simplify the fraction.

First, add up the whole amount of marbles.

3+5+2= 10

This is very simple, to find the probability, just put...

<u>number of blue marbles</u>

total number of marbles

This gives you, 5/10 and simplified into 1/2.

For the red marble, do the same thing.

<u>number of red marbles</u>

total number of marbles

This gives you, 3/10 and it can't be simplified.

A football team loses 5 2/3 yards on the first play. On the next play, they gain 2 3/4 yards. What is the overall increases or decreases in yards?

For this word problem, add 2 3/4 yards to 5 2/3 yards to find the overall decrease or increase in yards.

Solve the whole number parts..

5-2= 3

Solving the fractions part...

<u> 2 </u> - <u> 3 </u>= <u> 8 </u> - <u> 9 </u>=  <u>- 1 </u>

3     4     12    12     12

3- 1/12= 2 11/12

So, it was a 2 11/12 decrease.

3 0
3 years ago
What is the value of the expression below when c=5 and d=4 ?
Orlov [11]

Answer:

138

Step-by-step explanation:

6(52)−(5)(4)+8

=(6)(25)−(5)(4)+8

=150−(5)(4)+8

=150−20+8

=130+8

=138

8 0
3 years ago
Read 2 more answers
Milton spilled some ink on his homework paper. He can't read the coefficient of $x$, but he knows that the equation has two dist
Lera25 [3.4K]

Answer:

Sum = -81

Step-by-step explanation:

See the comment for complete question.

Given

c = 36 ----- Constant

No coefficient of x^2

Required:

Determine the sum of all distinct positive integers of the coefficient of x

Reading through the complete question, we can see that the question has 3 terms which are:

x^2 ---- with no coefficient

x ---- with an unknown coefficient

36 ---- constant

So, the equation can be represented as:

x^2 + ax + 36 = 0

Where a is the unknown coefficient

From the question, we understand that the equation has two negative integer solution. This can be represented as:

x = -\alpha and x = -\beta

Using the above roots, the equation can be represented as:

(x + \alpha)(x + \beta) = 0

Open brackets

x^2 + (\alpha + \beta)x + \alpha \beta = 0

To compare the above equation to x^2 + ax + 36 = 0, we have:

a = \alpha + \beta

\alpha \beta = 36

Where: \alpha, \beta and \alpha \ne \beta

The values of \alpha and \beta that satisfy \alpha \beta = 36 are:

\alpha = -1 and \beta = -36

\alpha = -2 and \beta = -18

\alpha = -3 and \beta = -12

\alpha = -4 and \beta = -9

So, the possible values of a are:

a = \alpha + \beta

When \alpha = -1 and \beta = -36

a = -1 - 36 = -37

When \alpha = -2 and \beta = -18

a = -2 - 18 = -20

When \alpha = -3 and \beta = -12

a = -3 - 12 = -15

When \alpha = -4 and \beta = -9

a = -4 - 9 = -13

At this point, we have established that the possible values of a are: -37, -20, -15 and -9.

The required sum is:

Sum = -37 -20 -15 - 9

Sum = -81

7 0
3 years ago
Which is the most appropriate to describe a quantity decreasing at a steady rate
frozen [14]
Graph C because it’s rate of change remains steadily constant as it decreases.
8 0
3 years ago
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The height h, in feet, of objects such as thrown balls can be modeled as a function of time t, in seconds, as h(t)=−16t2+v0t+h0,
Fed [463]

Answer:

Kerry's ball stays longer in the air than Marias ball by an extra 0.582 seconds

Step-by-step explanation:

The equation that models the height, h, in feet of object thrown modeled as a time function is given as follows;

h(t) = -16t² + v₀·t + h₀

Where;

v₀ = The velocity upwards (vertical velocity)

h₀ = The initial height of the object in feet

The given parameters are;

The initial height from which Maria throws a ball = 6 feet

The initial vertical velocity with which Maria throws the ball = 40 feet per second

The initial height from which Kerry throws a ball = 5 feet

The initial vertical velocity with which Kerry throws the ball = 50 feet per second

For Maria, from the equation for the height of the object, we have;

h(t) = -16t² + v₀·t + h₀

Where for Maria, we have;

h₀ = 6 feet

v₀ = 40 ft/s

Substituting gives;

h(t) = -16 × t² + 40 × t + 6 = -16·t² + 40·t + 6

h(t) = -16·t² + 40·t + 6

The time it takes for the ball to go up and come back down again is given by the value of t, when h = 0 (ground level) as follows;

0 = -16·t² + 40·t + 6

16·t² - 40·t - 6  = 0

From the quadratic formula, we have

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

t = \dfrac{40\pm \sqrt{(-40)^{2}-4\times 16 \times (-6)}}{2\times 16} = \dfrac{40 \pm 8 \times \sqrt{31} }{32}  = \dfrac{5}{4} \pm \dfrac{\sqrt{31} }{4}

t ≈ 2.64 s or -0.14 s

Therefore, the time the ball spends in the air ≈ 2.64 s

For Kerry,  we have;

h₀ = 5 feet

v₀ = 50 ft/s

Substituting gives;

h(t) = -16 × t² + 50 × t + 5 = -16·t² + 50·t + 5

h(t) = -16·t² + 50·t + 5

Which gives;

16·t² - 50·t - 5  = 0

From the quadratic formula, we have

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

t = \dfrac{50\pm \sqrt{(-50)^{2}-4\times 16 \times (-5)}}{2\times 16} = \dfrac{50 \pm 2 \times \sqrt{705} }{32}  = \dfrac{25}{16} \pm \dfrac{\sqrt{705} }{16}

t ≈ 3.222 s or -0.097 s

he time the ball spends in the air ≈ 3.222 s

Therefore, Kerry's ball that spends approximately 3.222 seconds stays longer in the air than Maria's ball that spends approximately 2.64 seconds in the air.

The difference in the two time duration is 3.222 - 2.64 = 0.582

Kerry's ball stays longer in the air than Marias ball by an extra 0.582 seconds.

4 0
3 years ago
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