Check the picture below.
is not very specific above, but sounds like it's asking for an equation for the trapezoid only, mind you, there are square tiles too.
but let's do the trapezoid area then,
![\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad \sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20a%5E%7B%5Cfrac%7B%7B%20n%7D%7D%7B%7B%20m%7D%7D%7D%20%5Cimplies%20%20%5Csqrt%5B%7B%20m%7D%5D%7Ba%5E%7B%20n%7D%7D%20%5Cqquad%20%5Cqquad%0A%5Csqrt%5B%7B%20m%7D%5D%7Ba%5E%7B%20n%7D%7D%5Cimplies%20a%5E%7B%5Cfrac%7B%7B%20n%7D%7D%7B%7B%20m%7D%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
34/100
= 0.34
3 tenths, 4 hundreths
Complete question is missing, so i have attached it.
Answer:
Percentile is 74th percentile
Step-by-step explanation:
All the lengths given are;
Bear Lengths 36.5 37.5 39.5 40.5 41.5 42.5 43.0 46.0 46.5 46.5 48.5 48.5 48.5 49.5 51.5 52.5 53.0 53.0 54.5 56.8 57.5 58.5 58.5 58.5 59.0 60.5 60.5 61.0 61.0 61.5 62.0 62.5 63.5 63.5 63.5 64.0 64.0 64.5 64.5 65.5 66.5 67.0 67.5 69.0 69.5 70.5 72.0 72.5 72.5 72.5 72.5 73.0 76.0 77.5
The number of lengths (inches) of bears given are 54 in number.
We are looking for the percentile corresponding to 65.5 in.
Looking at the lengths given, since they are already arranged from smallest to highest, let's locate the position of 65.5 in.
The position of 65.5 in is the 40th among 54 lengths given.
If the percentile is P, then;
P% x 54 = 40
P = (40 × 100)/54
P ≈ 74
Answer:
The correct option is (A) $304.47.
Step-by-step explanation:
The formula to compute the future value (<em>FV</em>) of an amount (A), compounded daily at an interest rate of <em>r</em>%, for a period of <em>n</em> years is:
![FV=A\times [1+\frac{r\%}{365}]^{n\times 365}](https://tex.z-dn.net/?f=FV%3DA%5Ctimes%20%5B1%2B%5Cfrac%7Br%5C%25%7D%7B365%7D%5D%5E%7Bn%5Ctimes%20365%7D)
The information provided is:
A = $300
r% = 1.48%
n = 1 year
Compute the future value as follows:
![FV=A\times [1+\frac{r\%}{365}]^{n\times 365}](https://tex.z-dn.net/?f=FV%3DA%5Ctimes%20%5B1%2B%5Cfrac%7Br%5C%25%7D%7B365%7D%5D%5E%7Bn%5Ctimes%20365%7D)
![=300\times [1+\frac{0.0148}{365}]^{365}\\\\=300\times (1.00004055)^{365}\\\\=300\times 1.014911\\\\=304.4733\\\\\approx \$304.47](https://tex.z-dn.net/?f=%3D300%5Ctimes%20%5B1%2B%5Cfrac%7B0.0148%7D%7B365%7D%5D%5E%7B365%7D%5C%5C%5C%5C%3D300%5Ctimes%20%281.00004055%29%5E%7B365%7D%5C%5C%5C%5C%3D300%5Ctimes%201.014911%5C%5C%5C%5C%3D304.4733%5C%5C%5C%5C%5Capprox%20%5C%24304.47)
Thus, the balance after 1 year is $304.47.
The correct option is (A).