A baseball is thrown upwards from a height of 5 feet with an initial speed of 64 feet per second, and its height h (in feet) fro
m the ground is given by h(t) = 5 + 64t – 16t2 where t is time in seconds. Using a graphing calculator, determine at what time the ball reaches its maximum height.
2 answers:
Answer:
2 seconds
Step-by-step explanation:
Given : A baseball is thrown upwards from a height of 5 feet with an initial speed of 64 feet per second, and its height h (in feet) from the ground is given by where t is time in seconds.
To Find: Using a graphing calculator, determine at what time the ball reaches its maximum height.
Solution:
Plot the graph of the given equation
Refer the attached graph
So, the given equation is a downward parabola
So, the y coordinate of vertex of the parabola will give its maximum height and x coordinate will given the time at which the ball is at maximum height
So, Vertex = (2,69)
Maximum height = 69 feet
Time to reach maximum height = 2 seconds
Hence it will take 2 seconds to reach its maximum height.
You might be interested in
Answer:
0.0901 = 9.01% of packages will incur such a charge
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions 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.
Sum of normal variables:
When we add normal variables, the mean will be the sum of the means, while the standard deviation is the square root of the sum of the variances.
Centerpieces are normally distributed with mean of 30 ounces and standard deviation of 2 ounces.
This means that
The weights of the shipping boxes are normally distributed with mean of 12 ounces and standard deviation of 1 ounce.
This means that
Suppose that centerpieces are chosen at random and packed into randomly chosen shipping boxes.
The shipment is given by the centerpice plus the shipping box. So the mean and the standard deviation of the weight are given by:
If a package exceeds 45 ounces, an additional charge is incurred. What is the proportion of packages will incur such a charge
This proportion is 1 subtracted by the pvalue of Z when X = 45. So
has a pvalue of 0.9099
1 - 0.9099 = 0.0901
0.0901 = 9.01% of packages will incur such a charge