Answer:
Cos C = adjecent/hypotenuse
Cos C = 14/50
divide the numerator and the denominator by 2
Cos C = 7/25
The answer is C
Answer:
The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
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 reading speed of a sixth-grader whose reading speed is at the 90th percentile
This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.




The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
Answer: it will take 125 additional minutes for both costs to be equal.
Step-by-step explanation:
Let a represent the number of additional minutes that it will it take for the two to be the same.
Company A charges a flat fee of 59.99 a month and .43 additional minutes. This means that the total cost of a additional units would be
0.43x + 59.99
Company B charges 69.99 a month and .35 for additional minutes. This means that the total cost of a additional units would be
0.35x + 69.99
For the cost of the two plans to be the same, it means that
0.43x + 59.99 = 0.35x + 69.99
0.43x - 0.35x = 69.99 - 59.99
0.08x = 10
x = 125
Switch the x and the y and solve for y
eg inverse of 2y= 3x+1
swith x and y
2x=3y+1
now solve for y
y =(2x-1)/3
The standard form of a circle is

where h and k are the center and x and y are the coordinates you're given. We need to solve for the radius to finish this off correctly. Filling in, we have

and

, giving us that

. Therefore, our equation is