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Vesna [10]
3 years ago
12

Suppose John opens a savings account with $1,000 that compounds interest daily. The APR at the time John deposits the account is

3.5%. He makes no withdrawals or deposits. What is his APY to the nearest hundredth of a percent after 1 year?
Mathematics
1 answer:
Fed [463]3 years ago
4 0
\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
\left. \qquad \qquad  \right.\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 3.5\%\to \frac{3.5}{100}\to &0.035\\
n=
\begin{array}{llll}
\textit{times it compounds per year}
\end{array}\to &365
\end{cases}
\\\\\\
\left(1+\frac{0.035}{365}\right)^{365}-1
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a construction company needs to remove 24 tons of dirt from a construction site. They can remove 3/8 tons of dirt each hour. How
Sunny_sXe [5.5K]
Since each hour is 3/8 of a tone then, namely how many times does 3/8 go into 24?  well, is just their quotient.

\bf 24\div\cfrac{3}{8}\implies \cfrac{24}{1}\div\cfrac{3}{8}\implies \cfrac{24}{1}\cdot\cfrac{8}{3}\implies \cfrac{8}{1}\cdot \cfrac{8}{1}\implies 64
4 0
3 years ago
Find the distance between -6,5 and -3,1
Elenna [48]

Answer:

11.7

Step-by-step explanation:

\sqrt{(-6-5)^2(-3-1)^2 }\\

=11.7

3 0
3 years ago
Toni wants to buy a shirt . The original price is $85 but it is on sale for 30% off Toni will pay15% sales tax how much will the
mixer [17]
 30%÷100% = 0.30; $85×0.30= $25.50; 85 - 25.50= $59.50.
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6 0
3 years ago
Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and
Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling:

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437

P(sample average sediment density is at most 3.00)

P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)

Calculation the value from standard normal z table, we have,  

P(x \leq 3.00) = 0.918

0.918 is the probability that the sample average sediment density is at most 3.00

4 0
3 years ago
True or false, there are two solutions to the equation |x-21|= -7
notsponge [240]

Answer:

false

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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