Answer:
113.04
Step-by-step explanation:
To find the area of a circle, you multiply 3.14 times the radius squared. To find the radius, you divided the diameter by 2. You get 6. You square 6 and get 36. Multiply that by 3.14 and you'll get 113.04 as your area.
keeping in mind that 4 months is not even a year, since there are 12 months in a year, 4 months is then 4/12 years.
let's assume is simple interest.
![\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$34300\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\to \frac{4}{12}\dotfill &\frac{1}{3} \end{cases} \\\\\\ A=34300\left[ 1+(0.035)\left( \frac{1}{3} \right) \right]\implies A= 34300(1.011\overline{6})\implies A=34700.1\overline{6}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%2434300%5C%5C%20r%3Drate%5Cto%203.5%5C%25%5Cto%20%5Cfrac%7B3.5%7D%7B100%7D%5Cdotfill%20%260.035%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B4%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B1%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D34300%5Cleft%5B%201%2B%280.035%29%5Cleft%28%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20A%3D%2034300%281.011%5Coverline%7B6%7D%29%5Cimplies%20A%3D34700.1%5Coverline%7B6%7D)
First we find the area of the two triangles on each side (keep in mind that they are equal/congruent to each other):
(24*10)/2 = 120 cm^2
Then you find the areas of the three rectangles that are left:
24*19 = 456 cm^2
26*19 = 494 cm^2
10*19 = 190 cm^2
Finally, to find the surface area of the prism:
120+120+456+494+190 = 1,380 cm^2
Answer:
The requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.
Step-by-step explanation:
A normal-distribution is an accurate symmetric-distribution of experimental data-values.
If we create a histogram on data-values that are normally distributed, the figure of columns form a symmetrical bell shape.
If X
N (µ, σ²), then
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z
N (0, 1).
The distribution of these z-variates is known as the standard normal distribution.
Thus, the requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.
Answer:
The attendance when the team won 4 games in the previous month is 525
Step-by-step explanation:
The attendance for a team's basketball game can be approximated with the polynomial : 
x represents the number of wins the team had in the previous month.
Factor the polynomial :





Substitute the value of x = 4
The attendance when the team won 4 games in the previous month = -5(4+3)(4-19)=525
Hence The attendance when the team won 4 games in the previous month is 525