$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$5000\\
P=\textit{original amount deposited}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, four times}
\end{array}\to &4\\
t=years\to &7
\end{cases}
\\\\\\
5000=P\left(1+\frac{0.05}{4}\right)^{4\cdot 7}\implies 5000=P(1.0125)^{28}\\\\\\ \cfrac{5000}{(1.0125)^{28}}=P$

8 0

2 years ago

An international polling agency estimates that 36 percent of adults from Country X were first married between the ages of 18 and

alukav5142 [94]

Answer:

(B) 0.2843

Step-by-step explanation:

Let d be the difference in proportions from Country X and Country Y who were first married between the ages of 18 and 32.

Then hypotheses are

$H_{0}$ : d=0.15

$H_{a}$ : d<0.15

Test statistic can be found using the equation

$z=\frac{p1-p2-0.15}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of Country X (0.36)

p2 is the sample proportion of Country Y (0.26)

p is the pool proportion of p1 and p2 ( $\frac{60*0.36+50*0.26}{50+60}=0.31$ )

n1 is the sample size of adults from Country X (60)

n2 is the sample size of adults from Country Y (50)

Then $z=\frac{0.36-0.26-0.15}{\sqrt{{0.31*0.69*(\frac{1}{60} +\frac{1}{50}) }}}$ ≈ 0.5646

p-value of test statistic is ≈ 0.2843

p-value states the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is at least 0.15

3 0

2 years ago