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The value of a $3000 computer decreases about 30% each year. write a function for the computers value V(t)

den301095 [7]

Answer:

Function for given situation is : $V(t)=3000(0.70)^t$

Value of computer after 4 years = $720.3.

Step-by-step explanation:

Given that the value of a $3000 computer decreases about 30% each year. Now we need to write a function for the computers value V(t). then we need to find about how much will the computer be worth in 4 years.

It clearly says that value decreases so that means function represents decay.

For decay we use formula:

$A=P(1-r)^t$

where P=initial value = $3000,

r= rate of decrease =30% = 0.30

t= number of years

A=V(t) = future value

so the required function is $V(t)=3000(1-0.30)^t$

or $V(t)=3000(0.70)^t$

Now plug t=4 years to get the value of computer after 4 years.

$V(4)=3000(0.70)^4$

$V(4)=720.3$

Hence final answer is $720.3.

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2 years ago

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can someone pls help me? i have no idea what to do. pls show the work steps for the questions. i will mark u as

tiny-mole [99]

Ok, so the question is based on geometric progression. Remember, the formula for calculating the nth-term of a geometric progression is: $a*r^{n-1}$ . The a in the expression stands for the 1st term of the sequence, and r is the common ratio of the elements of the sequence. Now let's take a look at the problem.

"A ping pong ball has a 75% rebound ration". We can infer that our common ratio, r, is 75% which is 0.75.

"When you drop it from a height of k feet...", this means the first height you drop it from, a.k.a, the first term.

Now going back to the expression, the nth-term = $a*r^{n-1}$ , we can substitute our common ration, 0.75 with r, and our 1st term, k, with a. This becomes: $k * 0.75^{n-1}$ . This becomes our expression.

a. The highest height achieved by the ball after six bounces. Our nth-term here is 6, so let's use our expression to find the 6th term. $n_{6} = 235 * 0.75^{6-1} = 235*0.75^{5} = 235*0.2373 = 55.7655ft$

b. The total distance travelled by the ball when it strikes the ground for the 12th time. This involves the use of the sum of elements in the geometric progression. The formula for that is $\frac{a(1 - r^{n})}{1-r}$ , provided that r is less than 1, which it is in this case, since 0.75 is less than one. Our nth-term here is 12, so we substitute.