Using the normal distribution, the areas to the left are given as follows:
a) 0.7910.
b) 0.6664.
c) 0.3707.
d) 0.8508.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X, and is also the area to the left of Z.
Hence:
- The area to the left of Z = 0.81 is of 0.7910.
- The area to the left of Z = 0.43 is of 0.6664.
- The area to the left of Z = -0.33 is of 0.3707.
- The area to the left of Z = 1.04 is of 0.8508.
More can be learned about the normal distribution at brainly.com/question/4079902
#SPJ1
6x^3 -24x^2 24x
3x^2 -12x 12
Answer:
#1 - Two complementary angles with measures that have a sum of 90.
#2- angles with a common vertex and side.
#3 - two supplementary angles with measures that have a sum of 180.
#4 - the region of a plane inside an angle
#5 - the region of a plane outside an angle
#6 - an angle with a measure of anything less than 90 degrees.
#7 - an angle with a measure of anything greater than 90 degrees.
These all should be right :)
Lmk how I did!
Answer:
The final velocity v is 21.74 m/s.
Step-by-step explanation:
Given:
![\textrm{initial velocity} = u = 9\ m/s\\\textrm{acceleration} = a = 7\ m/s^{2}\\\textrm{distance} = s = 28\ m](https://tex.z-dn.net/?f=%5Ctextrm%7Binitial%20velocity%7D%20%3D%20u%20%3D%209%5C%20m%2Fs%5C%5C%5Ctextrm%7Bacceleration%7D%20%3D%20a%20%3D%207%5C%20m%2Fs%5E%7B2%7D%5C%5C%5Ctextrm%7Bdistance%7D%20%3D%20s%20%3D%2028%5C%20m)
To find:
![\textrm{final velocity} = v = ?](https://tex.z-dn.net/?f=%5Ctextrm%7Bfinal%20velocity%7D%20%3D%20v%20%3D%20%3F)
Solution:
As we know the third equation of motion is represented as
![v^{2} = u^{2} + 2\times a\times s\\](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202%5Ctimes%20a%5Ctimes%20s%5C%5C)
Substituting the values u, a and s in the above equation we get
![v^{2} = 9^{2} + 2\times 7\times 28\\ =81 + 392\\=473\\](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%209%5E%7B2%7D%20%2B%202%5Ctimes%207%5Ctimes%2028%5C%5C%20%3D81%20%2B%20392%5C%5C%3D473%5C%5C)
![v =\±\sqrt{473} \\v =\sqrt{473} \ \textrm{as acceleration is positive v is also positive}\\v = 21.74\ m/s](https://tex.z-dn.net/?f=v%20%3D%5C%C2%B1%5Csqrt%7B473%7D%20%5C%5Cv%20%3D%5Csqrt%7B473%7D%20%5C%20%5Ctextrm%7Bas%20acceleration%20is%20positive%20v%20is%20also%20positive%7D%5C%5Cv%20%3D%2021.74%5C%20m%2Fs)