Answer:
108 miles
Step-by-step explanation:
Given the points (0.5, 135) and (2,81)
Diatnce between two points :
D = √(x2 - x1)² + (y2 - y1)²
X1 = 0.5 ; y1 = 135 ; x2 = 2 ; y2 = 81
D = √(2^2 - 0.5^2) + (81^2 - 135^2)
D = √(4 - 0.25) + (6561 - 18225)
D = √3.75 −11664
D = 107.98263
D = 108 miles
Answer:
69.14% probability that the diameter of a selected bearing is greater than 84 millimeters
Step-by-step explanation:
According to the Question,
Given That, The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 84 millimeters.
- In a set with mean and standard deviation, the Z score of a measure X is given by Z = (X-μ)/σ
we have μ=87 , σ=6 & X=84
- Find the probability that the diameter of a selected bearing is greater than 84 millimeters
This is 1 subtracted by the p-value of Z when X = 84.
So, Z = (84-87)/6
Z = -3/6
Z = -0.5 has a p-value of 0.30854.
⇒1 - 0.30854 = 0.69146
- 0.69146 = 69.14% probability that the diameter of a selected bearing is greater than 84 millimeters.
Note- (The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X)
Let's imagine we enclosed the entire shape in a 15 inch wide, 21 inch tall rectangle, area 315 sq inches.
From that rectangle we have to take away the 3x6 bottom left corner, and a strip 21-6=15 inches long and 3 inches wide on the right.
A = 15×21 - 3×6 - 3×15 = 315 - 63 = 252 sq in
Answer: D. 252 inches squared, last choice
