Elena has 14 pieces of bannana bread. She gives an equal amount of bannana bread to 4 friends. How many peices of bannana bread
2 answers:
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The given function is
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)
Answer:
a) 1186
b) Between 1031 and 1493.
c) 160
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 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.
Normally distributed with mean of 1262 and a standard deviation of 118.
This means that
a) Determine the 26th percentile for the number of chocolate chips in a bag.
This is X when Z has a p-value of 0.26, so X when Z = -0.643.
(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.
Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.
2.5th percentile:
X when Z has a p-value of 0.025, so X when Z = -1.96.
97.5th percentile:
X when Z has a p-value of 0.975, so X when Z = 1.96.
Between 1031 and 1493.
(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies?
Difference between the 75th percentile and the 25th percentile.
25th percentile:
X when Z has a p-value of 0.25, so X when Z = -0.675.
75th percentile:
X when Z has a p-value of 0.75, so X when Z = 0.675.
IQR:
1342 - 1182 = 160