Can someone solve number 22 step by step?

If $8259 principal is invested in a savings account that pays 5.25% interest compounded continuously how much money will be in t

VMariaS [17]

Answer:

A(10) = $13,961.50

Step-by-step explanation:

First, convert R as a percent to r as a decimal

r = R/100

r = 5.25/100

r = 0.0525 rate per year,

Then solve the equation for A

The formula is given as:

A = Pe^rt

P = 8259

r = 0.0525

t = 10 years.

Hence,

A = 8,259.00 × e^(0.0525×10)

A = $13,961.50

Therefore, the money that will be in the account after 10 years is $13,961.50

5 0

3 years ago

What value of w that makes the equation true 2/3(w+3) = 7

alexdok [17]

The answer is 7.5 or 15/2

3 0

3 years ago

12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests

Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that $p = 0.12$ .

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so $n = 18$