Please sleep me helpppp

Mathematics

1 answer:

defon2 years ago

5 0

Answer:

help me first on this question

Step-by-step explanation:

You might be interested in

Franklin deposits $3500 in an account that earns 3.5% interest compounded annually. What function equation represents the balanc

lara31 [8.8K]

So this problem can be solve using the formula:
F = P(1+i)^t
Where F is the balance account after t years
P is money deposited
i is the fraction interest rate
F = 3500 (1+0.35)^t
<span>F = 3500 (1.35)^t is the function equation represents the balance of the account after t years</span>

5 0

2 years ago

The area of a triangle is 150 cm^2. If the area of a triangle is given by the formula A=0/5b*h, write an equation which relates

taurus [48]

Answer:

Step-by-step explanation:

If the area of a triangle is 150cm^2, the formula would look like:

150=0.5b*h

So, if we were to solve for the base in terms of h, we'd simply divide both sides by 0.5b getting:

(150)/(0.5b)=h

which simplifies to

h=300b

4 0

3 years ago

Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

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

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have: