Umm maybe try:
f^-1 (x) = -4x/3 + 32/3
let me know if that’s right
First, put the numbers in order
A. 1,2,3,4,5,7,8,8,9.....median (middle number) is 5..when there is an odd number of data values, there will only be one median.
B. 14,14,15,16,17,18,19,20....if there is an even number of data points, there will be 2 middle numbers. In that case, u add the 2 middle numbers and divide by 2 to find the median. (16 + 17) / 2 = 33/2 = 16.5 is ur median.
Answer: your answer would be 5/16th :)
Answer:
a= 2
Step-by-step explanation:
2 x 2=4
7-4=3
Answer: 0.9649
Step-by-step explanation:
Let A denote the event that the days are cloudy and B denotes the event that the days are rainy.
Given : For the month of March in a certain city, the probability that days are cloudy :
Also in the month of March in the same city,, the probability that the days are cloudy and rainy :
Now by using the conditional probability, the probability that a randomly selected day in March will be rainy if it is cloudy will be :-

![\Rightarrow\ P(B|A)=\dfrac{0.55}{0.57}\\\\=0.964912280702\approx0.9649\ \ \text{[Rounded to four decimal places.]}](https://tex.z-dn.net/?f=%5CRightarrow%5C%20P%28B%7CA%29%3D%5Cdfrac%7B0.55%7D%7B0.57%7D%5C%5C%5C%5C%3D0.964912280702%5Capprox0.9649%5C%20%5C%20%5Ctext%7B%5BRounded%20to%20four%20decimal%20places.%5D%7D)
Hence, the probability that a randomly selected day in March will be rainy if it is cloudy = 0.9649