12/20=.6
Thus, move the decimal point two places to the right, and the answer is 60%
Use Ga_thMath(u) (brainly doesn't allow me to type it) To use the app u need to take a pic of the problem and then it will process it and you'll get ur answer ASAP(most of the time). Many questions have been asked before so search it on brainly.
Answer:
0.3333 (the three continues)
Step-by-step explanation:
If you put 1/3 or 1 divided by 3 in the calculator, then it'll show the decimal form of the fraction.
Answer:
Given definite integral as a limit of Riemann sums is:
![\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Step-by-step explanation:
Given definite integral is:
Substituting (2) in above
![f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=f%28x_%7Bi%7D%29%3D%5Cfrac%7B1%7D%7B2%7D%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%2B%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%5E%7B3%7D%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%282%2B%5Cfrac%7B3%7D%7B2n%7Di%29%2B%2864%2B%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B3%2816%29%5Cfrac%7B3%7D%7Bn%7Di%2B3%284%29%5Cfrac%7B9%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%29%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B3%7D%7B2n%7Di%2B%5Cfrac%7B144%7D%7Bn%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B291%7D%7B2n%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Riemann sum is:
![= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
So if we want to know the least amount, we first want to assume the other two games both had a score of 52, so we can say the last one had the least possible.
So if both got 52, the total points would be 104, for two games of 52 points. Since we want 141 points, we therefore want 37 more points to reach 141.
So the least amount of points a player could've scored in one of the games was 37.
