All categories

2 years ago

how much would 500 invested at 4% interest compounded continuously be worth after 7 years? round your answer to the nearest cent

Mathematics

1 answer:

vladimir1956 [14]2 years ago

4 0

Answer: the answer is 661.55

Step-by-step explanation:

interest formula: F = Pe^(rt)

You might be interested in

4/5 cup of chocolate chips is used to make 7 granola bars. How many granola bars can be made with 4 cups of chocolate chips?

s344n2d4d5 [400]

4/5=.80

4/.80= 5

5*7= 35

You can make 35 granola bars with 4 cups of chocolate chips.

I hope this helps!
~kaikers

8 0

3 years ago

Will give brainliest <br> Please answer and Explain detailed

AleksandrR [38]

Answer:

27%

Step-by-step explanation:

50+61=111

111÷4=27

4 is the number of roles

6 0

3 years ago

Read 2 more answers

If f(×) = -2x+8, find f(3)

andrezito [222]

Answer:

f(3) = 2

Step-by-step explanation:

To evaluate f(3), substitute x = 3 into f(x), that is

f(3) = - 2(3) + 8 = - 6 + 8 = 2

4 0

3 years ago

Find the value of x.<br><br> PLEASE HELP MEEEE!!! PLEASE AND THANK YOUU!!

Elenna [48]

The value of the x is 120

3 0

3 years ago

A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience

bulgar [2K]

Answer:

a) $P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

And replacing we got:

$P(X \geq 1) =1 -1x10^{-18} \approx 1$

b) $P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

c) $P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

Step-by-step explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

$X \sim Binom(n=9, p=0.99)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want to find this probability:

$P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

And replacing we got:

$P(X \geq 1) =1 -1x10^{-18} \approx 1$

Part b

$P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)$

$P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

Part c

$P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

4 0

3 years ago