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Yuliya22 [10]
3 years ago
15

How do I solve this ?

Mathematics
1 answer:
Burka [1]3 years ago
5 0
Your answer would be 
8y^7
-------
3X^3
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8000$ is invested in an account the yields 6% interest per year. After how many years will the account worth 13709.60$ if the in
ozzi

Answer:

After 9 years the account will be worth 13709.60$

Step-by-step explanation:

We are given the following in the question:

We are given the following in the question:

P = $8000

r = 6% = 0.046

n = 12

The compound interest is given by:

A = p\bigg(1+\dfrac{r}{n}\bigg)^{nt}

where A is the amount, p is the principal, r is the interest rate, t is the time in years.

Putting the values, we get,

13709.60 = 8000\bigg(1+\dfrac{0.06}{12}\bigg)^{12t}\\\\\dfrac{13709.60}{8000} = \bigg(1+\dfrac{0.06}{12}\bigg)^{12t}\\\\\Rightarrow 1.7137 = (1.005)^{12t}\\\Rightarrow t \approx 9

Thus, after 9 years the account will be worth 13709.60$

5 0
3 years ago
Read 2 more answers
Assume that in a statistics class the probability of receiving a grade of A equals .30 and the probability of receiving a grade
Tom [10]

Answer:

Answer D is correct

6 0
4 years ago
Write an equation in point-slope form for the line that has a slope of 4/5 and contains the point (−5, −3).
ankoles [38]

y = mx + b, where m = slope, and b = y-intercept.

Since you are not given the y-intercept, you have to solve for it through the equation, y = (4/5)x (which is the slope you are given) + b, and replace x and y with values given, which are (-5,-3)

y = (4/5)x + b

-3 = (4/5)(-5) + b

-3 = -4 + b

1 = b


Then replace b in the first equation to get the answer

y = (4/5)x + 1

5 0
3 years ago
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
The graph of y=4x has been shifted up 3 units.what is it’s new equation
Katyanochek1 [597]

I think it’s y=4x+3

4 0
3 years ago
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