Answer:
7.17 km
Step-by-step explanation:
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The length and width of a new rectangle playing field are 214 yards and 52 yards respectively.
<h3>What is the area of the rectangle?</h3>
It is defined as the area occupied by the rectangle in two-dimensional planner geometry.
The area of a rectangle can be calculated using the following formula:
Rectangle area = length x width
We have:
The length of a new rectangle playing field is 6 yards longer than quadruple the width.
Let's suppose the length is l and width is w of a rectangle:
From the problem:
l = 6 + 4w
Perimeter P = 2(l + w)
532 = 2(l + w)
Plug l = 6+4w in the above equation:
532 = 2(6 + 4w + w)
266 = 6 + 5w
260 = 5w
w = 52 yards
l = 6 +4(52) = 214 yards
Thus, the length and width of a new rectangle playing field are 214 yards and 52 yards respectively.
Learn more about the area of rectangle here:
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D.) Dependent variable is "<span>instantaneous velocity" as it gives information of velocity at every second
Hope this helps!</span>
Answer:
x = 4 ±
Step-by-step explanation:
Given
x² - 8x = 3
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(- 4)x + 16 = 3 + 16
(x - 4)² = 19 ( take the square root of both sides )
x - 4 = ± ( add 4 to both sides )
x = 4 ±
Answer:
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
Find the probability that the diameter of a selected bearing is greater than 85 millimeters.
This is 1 subtracted by the pvalue of Z when X = 85. Then
has a pvalue of 0.7486.
1 - 0.7486 = 0.2514
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.