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:

- 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
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Answer:
1.5 unit^2
Step-by-step explanation:
Solution:-
- A graphing utility was used to plot the following equations:

- The plot is given in the document attached.
- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).
- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.
We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).
The double integration formulation can be written as:

Answer: 1.5 unit^2 is the amount of area bounded by the given curve f ( x ) and the boundary equations.
Answer:
x= 2.28
or, x= 0.219
Step-by-step explanation:
2x²-5x+1 =0
a= 2, b= -5 and c= 1
x= (-b <u>+</u> √(b²-4ac))/2a
= (5 <u>+</u> √(25-8))/4
= (5 <u>+</u> √17)/4
Each side is 6 feet.
let s be 1 side of the square
24 = 4s
6 = s