All categories

3 years ago

Nolan deposited $2300 into a savings account for which interest is compound weekly at a rate of 2.12%.

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

7 0

Answer:

The answers is <u><em>367.87</em></u>

Step-by-step explanation:

Shkiper50 [21]3 years ago

6 0

p(1+r/n)^(r times n)

P is your start amount

R is rate as a decimal 5% would be 0.05

t is time (years)

n is the number of time compounded , weekly would be 52 times a year.

2300(1 + (0.0212/52))^(52 times 7)

2300(1.0004076)^(364)

You might be interested in

Wanda opened a savings account 20 years ago with a deposit of $3,150.37. The account has an interest rate of 4.1% compounded dai

Aleksandr [31]

$\bf ~~~~~~ \stackrel{daily}{\textit{Continuously Compounding Interest Earned Amount}}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$3150.37\\
r=rate\to 4.1\%\to \frac{4.1}{100}\to &0.041\\
t=years\to &20
\end{cases}
\\\\\\
A=3150.37e^{0.041\cdot 20}\implies A=3150.37e^{0.82}\implies A\approx 7152.91457$

how much interest is there? well, A - 3150.37.

3 0

3 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448$

$12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152$

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0

3 years ago

What is the answer!?!?!?

77julia77 [94]

The price for hardcover books is 1.50 and the price for paperback is .50

8 0

3 years ago

Write an expression for the sequence of operations described below.

den301095 [7]

Answer:

We conclude that:

''add 3 and the sum of 9 and v'' is algebraically represented by the expression as:

3 + 9 + v

Step-by-step explanation:

Given the statement

''add 3 and the sum of 9 and v''

Let us break down the statement

Let the number be: v

The sum of 9 and v: 9+v

Adding 3 and the sum of 9 and v will be: 3 + 9 + v

Therefore, we conclude that:

''add 3 and the sum of 9 and v'' is algebraically represented by the expression as:

3 + 9 + v

3 0

3 years ago

-1/4x+2y=28 find the x and y intercept

Advocard [28]

The y intercept is (0, 14) and the x intercept is (-112,0) all you have to do is just make y zero if you are trying to solve for x and the same for the x if you are trying to solve for y.

8 0

3 years ago