Kl =4x+3 jk =4x+2 and jl =45 find kl

If you need $8000 to travel to Europe after you graduate in 4 year. How much would your monthly deposits need to be if your acco

lina2011 [118]

$\bf \qquad \qquad \textit{Future Value of an ordinary annuity}\\
\left. \qquad \qquad \right.(\textit{payments at the end of the period})
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]$

$\bf \qquad 
\begin{cases}
A=
\begin{array}{llll}
\textit{accumulated amount}\\
\end{array}\to &
\begin{array}{llll}
8000
\end{array}\\
pymnt=\textit{periodic payments}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12\\
t=years\to &4
\end{cases}$

$\bf 8000=pymnt\left[ \cfrac{\left( 1+\frac{0.05}{12} \right)^{12\cdot 4}-1}{\frac{0.05}{12}} \right]
\\\\\\
\cfrac{8000}{\left[ \frac{\left( 1+\frac{0.05}{12} \right)^{12\cdot 4}-1}{\frac{0.05}{12}} \right]}=pymnt\implies \cfrac{8000}{\frac{\left( \frac{241}{240} \right)^{48}-1}{\frac{1}{240}}}=pymnt
\\\\\\
\cfrac{8000}{53.0148852}\approx pymnt\implies 150.9010152318\approx pymnt$

6 0

3 years ago

The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{670 - 660}{9}$

$Z = 1.11$

$Z = 1.11$ has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0

2 years ago

Which solution is correct? Be sure to check for extraneous solutions.

denis23 [38]

That answer is d)x=-4

4 0

3 years ago

If you invested $2000 in an account that paid 8% annual interest compounded quarterly, how much would the account be worth in 10

Morgarella [4.7K]

Answer:

4,317.85

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

after an art class 15% of the color pencils were missing. how many color pencils were there originally if 9 color pencils are no

NARA [144]

We know that the 15% of missing pencils is equal to 9

we can use this formula

9/x = 15/100

Then cross multiply

15x = 900

x = 60

There were orignally 60 pencils

8 0

3 years ago