Sum of two prime numbers = 85
40+45=85
Sum of two product numbers = 40*45
=1800
Hope I helped
The length and the width of the rectangle are 22 inches and 16 inches respectively
<h3>How to determine the length and the width of the rectangle?</h3>
The given parameters are:
Length = 6 inches + width
Perimeter = 76 inches
The perimeter is calculated as:
Perimeter = 2 * (Length + Width)
This gives
76 inches = 2 * (6 inches + Width + Width)
Divide through by 2
38 inches = 6 inches + Width + Width
Evaluate the like terms
2 * Width = 32 inches
Divide both sides by 2
Width = 16 inches
Substitute Width = 16 inches in Length = 6 inches + width
Length = 6 inches + 16 inches
Evaluate
Length = 22 inches
Hence, the length and the width of the rectangle are 22 inches and 16 inches respectively
Read more about perimeters at:
brainly.com/question/24571594
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The graph is shown in the attached image.
Answer:
The standard deviation of given probability distribution is 0.767.
Step-by-step explanation:
We are given the following information in the question:
X: 0 1 2 3 4
P(x): 0.4521 0.3970 0.1307 0.0191 0.0010
Formula:
![\text{Mean} = \sum X.P(x)\\= 0(0.4521) + 1(0.3970) + 2(0.1307) + 3(0.0191) + 4(0.0010)\\= 0.7197](https://tex.z-dn.net/?f=%5Ctext%7BMean%7D%20%3D%20%5Csum%20X.P%28x%29%5C%5C%3D%200%280.4521%29%20%2B%201%280.3970%29%20%2B%202%280.1307%29%20%2B%203%280.0191%29%20%2B%204%280.0010%29%5C%5C%3D%200.7197)
![\mu = 0.7197](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.7197)
Formula:
![\text{Variance} = \sum (X-\mu)^2E(x)\\= (0-0.7197)^2(0.4521)+(1-0.7197)^2(0.3970)+(2-0.7197)^2(0.1307)+(3-0.7197)^2(0.0191)+(4-0.7197)^2(0.0010) \\=0.587](https://tex.z-dn.net/?f=%5Ctext%7BVariance%7D%20%3D%20%5Csum%20%28X-%5Cmu%29%5E2E%28x%29%5C%5C%3D%20%280-0.7197%29%5E2%280.4521%29%2B%281-0.7197%29%5E2%280.3970%29%2B%282-0.7197%29%5E2%280.1307%29%2B%283-0.7197%29%5E2%280.0191%29%2B%284-0.7197%29%5E2%280.0010%29%20%5C%5C%3D0.587%20)
![\text{Standard Deviation} = \sqrt{\text{Variance}}\\= \sqrt{0.587} = 0.767](https://tex.z-dn.net/?f=%5Ctext%7BStandard%20Deviation%7D%20%3D%20%5Csqrt%7B%5Ctext%7BVariance%7D%7D%5C%5C%3D%20%5Csqrt%7B0.587%7D%20%3D%200.767)
The standard deviation of given probability distribution is 0.767.