Multiplying both sides of the equation by 3 and substituting 4 for p
Explanation:
The easiest way to do this is to make use of the 2-point form of the equation for a line. For points (x₁, y₁) and (x₂, y₂), the equation is ...
![y=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%28x-x_1%29%2By_1)
Filling in your given points, the equation becomes ...
![y=\dfrac{7-1}{\frac{-7}{2}-\frac{9}{2}}(x-\frac{9}{2})+1](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B7-1%7D%7B%5Cfrac%7B-7%7D%7B2%7D-%5Cfrac%7B9%7D%7B2%7D%7D%28x-%5Cfrac%7B9%7D%7B2%7D%29%2B1)
After you fill in the values, it is a matter of simplifying the resulting equation.
![y=\dfrac{6}{\frac{-16}{2}}(x-\frac{9}{2})+1=\dfrac{-12}{16}(x-\frac{9}{2})+1=-\dfrac{3}{4}x+\left(\dfrac{-3}{4}\cdot\dfrac{-9}{2}\right)+1\\\\y=-\dfrac{3}{4}x+\dfrac{27+8}{8}\\\\y=-\dfrac{3}{4}x+\dfrac{35}{8}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B6%7D%7B%5Cfrac%7B-16%7D%7B2%7D%7D%28x-%5Cfrac%7B9%7D%7B2%7D%29%2B1%3D%5Cdfrac%7B-12%7D%7B16%7D%28x-%5Cfrac%7B9%7D%7B2%7D%29%2B1%3D-%5Cdfrac%7B3%7D%7B4%7Dx%2B%5Cleft%28%5Cdfrac%7B-3%7D%7B4%7D%5Ccdot%5Cdfrac%7B-9%7D%7B2%7D%5Cright%29%2B1%5C%5C%5C%5Cy%3D-%5Cdfrac%7B3%7D%7B4%7Dx%2B%5Cdfrac%7B27%2B8%7D%7B8%7D%5C%5C%5C%5Cy%3D-%5Cdfrac%7B3%7D%7B4%7Dx%2B%5Cdfrac%7B35%7D%7B8%7D)
Answer:
if it is it parenthesis it means to multiply. Every equation has to have an symbol all symbols are different but the same. If that equation have an letter you multiply that number times the letter.
Answer:
Rs 1155.3
Step-by-step explanation:
1136 * .1 = 113.6
1136 - 113.6 = 1022.4
1022.4 * 1.13 = 1155.3
Answer:
A) 0.335; B) 0.343; C) 0.599
Step-by-step explanation:
For part A,
To find the probability that the user has more than one infection per year, we can either add together the probabilities for 2, 3, 4 and 5; or we can add together the probabilities for 0 and 1 and subtract them from 1:
1-(P(X = 0)+P(X = 1))
= 1-(0.343+0.322) = 1-0.665 = 0.335
For part B,
The probability that the user has no infections per year is P(X = 0); this is 0.343.
For part C,
The probability that the user has between 1 and 3 infections (inclusive) per year is
P(1 ≤ X ≤ 3) = 0.322+0.201+0.076 = 0.599