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Tpy6a [65]
2 years ago
15

Ashley invests $9,720 in a one-month money market account paying 3.16% simple annual interest and $8,140 in a two-year CD yieldi

ng 3.23% simple annual interest. Assuming Ashley does not reinvest or renew these investments, how much money will she have when both investments reach maturity, to the nearest dollar?

Mathematics
2 answers:
tester [92]2 years ago
5 0
For the first investment. A = P(1 + rt); where p = 9,720, r = 0.0316 and t = 1/12
A = 9720(1 + 0.0316/12) = 9720(1.0026) = $9,746

For the second investment,
A = 8140(1 + 0.0323 x 2) = 8140(1.0646) = $8,666

Total amount she had = $9,746 + $8,666 = $18,412
Elodia [21]2 years ago
3 0

Answer:

D. $18,411

Step-by-step explanation:

I just answered it correct

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How much more would $1000 earn in 5 years in an account compounded continuously than an account compounded quarterly if the inte
choli [55]
\bf ~~~~~~ \textit{Compounding Continuously Interest Earned Amount}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$1000\\
r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\
t=years\to &5
\end{cases}
\\\\\\
A=1000e^{0.037\cdot 5}\implies A=1000e^{0.185}\\\\
-------------------------------\\\\

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &5
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.037}{4}\right)^{4\cdot 5}\implies A=1000(1.00925)^{20}

compare both amounts.
8 0
2 years ago
At a snack stand,drinks are $1.50. Write an expression that could be used to find the total cost in dollars of d drinks?
Mashcka [7]
1.50d is the right answer since we don't know how many drinks there are
4 0
1 year ago
There are approximately 0.4536 kilograms in a pound. To the nearest whole pound, how many pounds does a steer that weighs 200 kg
Sergio039 [100]

Answer:

i need help with this one too plz help

Step-by-step explanation:

5 0
1 year ago
Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents wh
kherson [118]

Answer:

a. The alternative hypothesis H₀: p'₁ ≠ p'₂ is accepted

b. Type I error

Step-by-step explanation:

Proportion of California residents who reported insufficient rest = 8.0%

Proportion of Oregon  residents who reported insufficient rest = 8.8%

p'₁ = 0.08 * 11545 =923.6

p'₂ = 0.088 * 4691=412.81

σ₁ = \sqrt{n*p_1*q_1}  = \sqrt{n*p_1*(1-p_1)} = \sqrt{11545*0.08*(1-0.08)} = 29.15

σ₂ = \sqrt{n*p_2*q_2}  = \sqrt{n*p_2*(1-p_2)}= \sqrt{4691*0.088*(1-0.088)} = 19.40

Samples size of California residents n₁ = 11,545

Samples size of Oregon residents n₂ = 4,691

Hypothesis can be constructed thus

Let our null hypothesis be H ₀: p'₁ = p'₂

and alternative hypothesis H ₐ: p'₁ ≠ p'₂

Then we have  

z =\frac{(p'_1 -p'_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2} } }

The test statistics can be computed by

     

t₀ = \sqrt{\frac{n_1n_2(n_1+n_2-2)}{n_1+n_2} } *\frac{p_1'-p_2'}{\sqrt{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2} } =      1104.83

c from tables is   P(T ≤ c) = 1 - α where α = 5% and c = 1.65

since t₀ ≥ c then then the hypothesis is rejected which means the alternative hypothesis is rejected

b. Type I error, rejecting a true hypothesis

5 0
2 years ago
The ratio of the number of beads Karen had to the number of beads Patricia had was 2:5 .After Patricia bought another 75 beads,t
blondinia [14]
K:P=2:5
P+75 it is 4:15
at first, P=5 units
so we have
2:5 and 4:15
since the K units didn't change we conver them to the same
2:5 times 2:2=4:10
so from 4:10 to 4:15 is a change of 75 in the right side
therefor 75=5 units so 75/5=15/1= 1 unit
so at first K=4 units=4 times 1 unit=4 times 15=60
P=10 units so 10 times 15=150

at first
Karen=60
Patricia=150
7 0
2 years ago
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