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2 years ago

if 5000000 invested at 4% interest compounded continuously how much will the investment be worth in 30 years

2 answers:

8 0

For this you use A=Pe^(rt)
A=furutre amount
P=present amount
e=natural base
r=rate in decimal
t=times in years

A=?
P=5000000
r=0.04
t=30

A=5,000,000e^(0.04*30)
A=5,000,000e^(1.2)
A=16600584.613683

Anna35 [415]2 years ago

6 0

The required formula is:
$A=Pe^{rt}$
where A is the amount after t years, e is the exponential number and r is the annual interest rate expressed as a decimal:
Plugging in the given values into the formula:
$A=5000000\times e^{(0.04\times 30)}=16,600,584.61\ currency\ units$

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Which of the following groups has terms that can be used interchangeably with the others?1. Percentage, Probability, and Proport

Dafna1 [17]

Answer: 1) Percentage, Probability, Proportion

Step-by-step explanation:

In statistics,

▪Proportion means a fraction of the total.

▪Percentage means a number or a ratio, expressed as a fraction of 100. Here, the total is 100.

▪Probability which may look slightly different from the other two means a number between 0 and 1, showing the exact likelihood of an event happening. Which in context can mean, a number showing a fraction of an event happening out of the whole (0 to 1).

▪Critical value means a point in hypothesis testing on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis.

In the four definitions, the odd one out is Critical Value.

So the option without critical value in it is option 1) Percentage, Probability, Proportion since we can use the three interchangeable.

6 0

2 years ago

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Determine the equation of the line that passes through the given points (if you have a graphing calculator, you can us the table

vladimir2022 [97]

Answer:

d. y = -2×

Step-by-step explanation:

First to solve find the slope using the formula, $\frac{10+10}{-5-5}$ . The slope equals 20/-10 or -2. Then plug the slope and point, (-5,10), into slope-point form. y-10=-2(x+5), then solve for y. This gives you y=-2x+0 or y=-2x.

8 0

1 year ago

A high school coach needs to buy new athletic shorts for the 15 members of the basketball team. The coach must spend less than $

Studentka2010 [4]

Answer:

15x < 200; x < 13.33; the maximum price for a pair of shorts

Step-by-step explanation:

1. Set up the inequality

Let x = price of a pair of shorts. Then

15x = price of shorts for the team

You have one condition:

15x < 200

2. Solve the inequality

$\begin{array}{rcl}15x & < & 200\\\\x & < & \dfrac{200}{15}\\\\x & < & \mathbf{13.333}\\\end{array}$

3. Meaning of solution

The solution represents the maximum price the coach can pay for a pair of shorts.

If the coach pays $13.33 per pair, the total cost for the team will be $199.95, and the condition is satisfied.

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2 years ago

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Given two terms in a geometric sequence, find the 8th term and the recursive formula.

Shtirlitz [24]

Answer:

Step-by-step explanation:

Finding the 8th term of this sequence will be much easier with an explicit formula, so we will find that, find the 8th term, then write the equation for the recursive. The recursive formula is only good if you know all the terms before the one you're looking for. That means if we want the 8th term, we first have to find the 7th term, but before that we have to find the 6th term, etc...

We can find the explicit formula for this using the info we have about the 5th term and the 3rd term. The explicit formula looks like this, in general terms:

$a_n=a_1*r^{n-1$ where a1 is the first term and r is the common ratio. We can solve for both, one at a time. First, using the fact that the 5th term is 3888:

$a_5=a_1*r^{5-1$ and subbing in and simplifying:

$3888=a_1*r^4$ and we will solve that for a1:

$a_1=\frac{3888}{r^4}$

Now the other equation we need for this system, using the fact that the third term is 108:

$108=a_1*r^2$ and we sub in our expression for a1:

$108=(\frac{3888}{r^4})*r^2$ which simplifies to

$108=\frac{3888}{r^2}$ and

$r^2=\frac{3888}{108}$ so

r = 6. Now we'll plug 6 in for r to find a1:

$a_1=\frac{3888}{6^4}$ so

a1 = 3. Now we have the a and r for the explicit formula:

$a_n=3*6^{n-1$ and we simply put in an for n:

$a_8=3*6^{8-1$ and

$a_8=3*6^7$ which gives us the 8th term as

a8 = 839808

The recursive formula is

$a_n=a_{n-1}*r$ and it works, also, but again, you have to have all the terms that come before the one you want or this formula is useless. Explicit is always better, in my opinion.