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user100 [1]
3 years ago
7

A bank charges a fee of 0.5% per month for having a checking account. Stephani’s account has $325 in it. Which function models t

he balance of Stephani’s account, B(t), in dollars, as a function of time, t, in months?
a: B(t) = 325(0.0995)t

b: B(t) = 325(0.005)t

c: B(t) = 0.05(325)t

d: B(t) = 325 + 12(0.005)t
Mathematics
1 answer:
Alex_Xolod [135]3 years ago
5 0

Answer:

f(t) = 325(0.995)^t

Step-by-step explanation:

Given

r= 0.5\% ---- Charges

a= \$325 --- Amount

Required

Represent the balance as a function

This is calculated using exponential function.

f(t) = ab^t

Where

b = 1 - r

So, we have:

b = 1 - 0.5\%

b = 1 - 0.005

b = 0.995

So:

f(t) = ab^t

f(t) = 325 * 0.995^t

f(t) = 325(0.995)^t

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What is the surface area of the cylinder
Hitman42 [59]

Answer:

In terms of pi: 164.5pi cm squared

Substituting pi for 3.14: 516.53 cm squared

Step-by-step explanation:

3.5^2 = 12.25pi

12.25*2 = 24.5pi

7pi*20 = 140pi

140 + 24.5 = 164.5pi

164.5*3.14 = 516.53

5 0
2 years ago
Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat
drek231 [11]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

Step-by-step explanation:

For this case we have the following probability distribution given:

X          0            1        2         3        4         5

P(X)   0.031   0.156  0.313  0.313  0.156  0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

We can verify that:

\sum_{i=1}^n P(X_i) = 1

And P(X_i) \geq 0, \forall x_i

So then we have a probability distribution

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

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It would be 18$ because you add the 8 plus the 10
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Read 2 more answers
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