0.074n = 74
multiply both sides by 1000 to make the process simpler
74n = 74000
divide both sides by 74
n = 1000
Answer:
See attached image for the drawing of the first four trees (circled in green)
The patterns is:
x = 2+3n and y= 3+2n
Position of 7th tree is: (20,15) (circled in orange in the image)
Step-by-step explanation:
Starting at the location (2,3) the next x and y positions are given by:
x = 2+3n since the horizontal position needs to be increased by 3 units on each iteration,
and y= 3+2n since the vertical position needs to be increased by 2 units on each iteration
being n= 1 through 6 (to account for the next 6 trees that need to be planted)
With such pattern, the location of the seventh tree would be:
x = 2 + 3*6 =2 + 18 = 20
y = 3 + 2*6 = 3 + 12 = 15
That is, the point (20,15) on the plane.
Also see attached image.
Answer:
The first digit of the quotient should be placed at the leftmost place of the places of the all the digits in the quotient.This is so from the basic rule of division.
Step-by-step explanation:
The quotient is given by,
[where [x] is the greatest integer function on x]
= [322.6]
= 322
and the remainder is given by,
= 9
So, the first digit of the quotient should be placed at the leftmost place of the places of the all the digits in the quotient and this is so from the very basic rule of division.
Answer:
Exterior angle = sum of opposite interior angles
Step-by-step explanation:
Answer:
7.3% of the bearings produced will not be acceptable
Step-by-step explanation:
Problems of normally distributed samples can be 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:
Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.
So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.
Larger than 0.504
1 subtracted by the pvalue of Z when X = 0.504.
has a pvalue of 0.9938
1 - 0.9938= 0.0062
Smaller than 0.496
pvalue of Z when X = -1.5
has a pvalue of 0.0668
0.0668 + 0.0062 = 0.073
7.3% of the bearings produced will not be acceptable