Which conversion factors are used to multiply to 18 cm/s18 cm/s to get meters per minute?
1 answer:
You might be interested in
Answer:
23 years
Step-by-step explanation:
Step 1: Calculate the rate constant (k) for the radioactive decay
A radioactive substance with initial concentration [A]₀ decays to 97% of its initial amount, that is, [A] = 0.97 [A]₀, after t = 1 year. Considering first-order kinetics, we can calculate the rate constant using the following expression.
ln [A]/[A]₀ = - k.t
k = ln [A]/[A]₀ / -t
k = ln 0.97 [A]₀/[A]₀ / -1 year
k = 0.03 year⁻¹
Step 2: Calculate the half-life of the substance
We will use the following expression.
= ln2/ k = ln 2 / 0.03 year⁻¹ = 23 years
Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ± *
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71
[10.41±1.645*]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!
Check the picture below.
so the chord is the one in red there, making the segment, in green there, with a central angle of 45°, and the minor segment will be that green one.
![\bf \textit{area of a segment of a circle}\\\\ A=\cfrac{r^2}{2}\left[ \cfrac{\pi \theta }{180} - sin(\theta) \right]~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=3\\ \theta =45 \end{cases} \\\\\\ A=\cfrac{3^2}{2}\left[ \cfrac{\pi(45) }{180} - sin(45^o) \right]\implies A=\cfrac{9}{2}\left[ \cfrac{\pi }{4}-\cfrac{\sqrt{2}}{2} \right] \\\\\\ A=\cfrac{9}{2}\left( \cfrac{\pi -2\sqrt{2}}{4} \right)\implies A\approx 0.3523112199491](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5C%0AA%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%5B%20%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D%20-%20sin%28%5Ctheta%29%20%5Cright%5D~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A%5Ctheta%20%3Dangle~in%5C%5C%0A%5Cqquad%20degrees%5C%5C%0A------%5C%5C%0Ar%3D3%5C%5C%0A%5Ctheta%20%3D45%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D%5Ccfrac%7B3%5E2%7D%7B2%7D%5Cleft%5B%20%5Ccfrac%7B%5Cpi%2845%29%20%7D%7B180%7D%20-%20sin%2845%5Eo%29%20%5Cright%5D%5Cimplies%20A%3D%5Ccfrac%7B9%7D%7B2%7D%5Cleft%5B%20%5Ccfrac%7B%5Cpi%20%7D%7B4%7D-%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0AA%3D%5Ccfrac%7B9%7D%7B2%7D%5Cleft%28%20%5Ccfrac%7B%5Cpi%20-2%5Csqrt%7B2%7D%7D%7B4%7D%20%5Cright%29%5Cimplies%20A%5Capprox%200.3523112199491)
so the chord is the one in red there, making the segment, in green there, with a central angle of 45°, and the minor segment will be that green one.