Answer:
104.127 m
Step-by-step explanation
In the question instead of chi, it would be kite.
Given: Distance of Brooke from the base of tree is 50 meters.
The kite is directly above the 10 m tree in the 125 m string is fully extended.
Lets assume the height of the kite be "x".
Its making the Right angle triangle with length of string as hypotenuse, height of kite above tree.
Using Pythagorean formula:
h²= a²+b²
⇒
⇒
⇒ 
Subtracting both side by 2500
⇒ 
Square rooting both side, we know√a²=a
⇒ 
∴ x= 114.127 meter.
Next, we know the value of x include the height of tree.
∴ Height of kite flying above tree= 
⇒ Height of kite flying above tree= 
Hence, the height of kite above the tree is 104.127 meters.
D(t) = -4t^3 + 40t^2 + 500t = 4088
4t^3 - 40t^2 - 500t + 4088 = 0
t = 7
Therefore, the plane has flown 4088 km 7 hours after it took off.
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.
A) 1.75p + 12 = 54
1.75p = 42
p = 24 bags
b) 1.75(70) + 12 = b
122.5 + 12 = b
b = $134.50
