Solve the system using Elimination
2 answers:
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Step-by-step explanation:
Given
To get g(x), we will have to integrate g'(x)
If g(1) = 0, this means at x = 1, g(x) = 0
0 = -1⁻¹ + C
C= 1
Substitute C = 1 into the function
g(x) = -x⁻¹ + 1
If g(2) = 0, this means at x = 2, g(x) = 0
0 = -2⁻¹ + C
C= 2⁻¹
C = 1/2
Substitute C = 2 into the function
g(x) = -x⁻¹ + 1/2
Recall your d = rt, distance = rate * time.
the runner takes off and goes at 4mph.
the cyclist takes off 2 hours later, and goes 14mph.
now, when they both meet, namely the cyclist comes from behind and meets ahead the runner, the distances both of them travelled, is say "d" miles for both, since both of them are "d" miles from the starting point.
if by them the cyclist has been going for say "t" hours, we know she took 2 hours later, so by the the runner has been running for "t + 2" hours then.
the runner takes off and goes at 4mph.
the cyclist takes off 2 hours later, and goes 14mph.
now, when they both meet, namely the cyclist comes from behind and meets ahead the runner, the distances both of them travelled, is say "d" miles for both, since both of them are "d" miles from the starting point.
if by them the cyclist has been going for say "t" hours, we know she took 2 hours later, so by the the runner has been running for "t + 2" hours then.
Answer:
a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.
b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.
c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute
Step-by-step explanation:
Empirical Rule:
The Empirical Rule states that, for a normally distributed random variable:
Approximately 68% of the measures are within 1 standard deviation of the mean.
Approximately 95% of the measures are within 2 standard deviations of the mean.
Approximately 99.7% of the measures are within 3 standard deviations of the mean.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean:
(95% of data) range from 19 to 75 beats per minute.
This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So
a. What is the probability that the heart rate is less than 25 beats per minute?
This is the p-value of Z when X = 25. So
has a p-value of 0.0582.
0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.
b. What is the probability that the heart rate is greater than 60 beats per minute?
This is 1 subtracted by the p-value of Z when X = 60. So
has a p-value of 0.8238.
1 - 0.8238 = 0.1762
0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.
c. What is the probability that the heart rate is between 25 and 60 beats per minute?
This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So
0.8238 - 0.0582 = 0.7656
0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute
This should help you out!
