Circumference of Circle A is is 42pi.
It is the same for all circles. To find the circumference, one must use the constant equation of 2pir. This equation stays the same, no matter what.
(6.2) and the perpendicularly to y = to 1223
Answer:
a = 2.5
b = 4.33
Step-by-step explanation:
a = sin(30 degrees) = o/5 = 2.5
b = cos(30 degrees) = a/5 = (5root3) / 2 or about 4.33
verify using Pythagorean theorem
2.5^2 + 4.33^2 = 5^2
6.25 + 18.75 = 25
25 = 25
Answer:
A(r) = √2 * r
A(r) Domain is R { r ; r > 0}
Step-by-step explanation:
Diagonals of a square intercept each other in a 90° angle. The four triangles resulting from diagonal interception are equal and are isosceles triangles, with hipotenuse a side of the square
Therefore we apply Pythagoras theorem
Let x be side of square, and r radius of the circle, ( diagonals touch the circle) then
x² = r² + r²
x² = 2r²
x = √2 * r
Now Aea of square is :
A = L² where L is square side
A(r) = √2 * r
Domain of A(r) = R { r, r > 0}
Answer:
The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours
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 problem, we have that:

What is the minimum level for which the battery pack will be classified as highly sought-after class
At least the 100-10 = 90th percentile, which is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.




The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours