2. Sarah has some blue marbles, twice as many red marbles as blue, and 8 yellow marbles. If she has a total of 32 marbles, how
1 answer:
Given:
Number of blue marbles, twice as many red marbles as blue, and 8 yellow marbles.
She has a total of 32 marbles.
To find:
The number of red marbles.
Solution:
Let number of blue marbles be x.
Then, number of red marbles = 2x
Number of yellow marbles = 8
Total number of marbles = 32
Now,
Number of red marbles
Therefore, there are 16 red marbles.
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Answer:
a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.
b) The weight that 80% of the apples exceed is of 78.28g.
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.
Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.
This means that
a. Find the probability a randomly chosen apple exceeds 100 g in weight.
This is 1 subtracted by the p-value of Z when X = 100. So
has a p-value of 0.9697
1 - 0.9696 = 0.0304
0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.
b. What weight do 80% of the apples exceed?
This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.
The weight that 80% of the apples exceed is of 78.28g.
Answer:
Step-by-step explanation:
Given:
The equation of the known line is:
A point on the unknown line is (-4, 4)
Now, since the two lines are parallel, their slopes must be equal.
Now, slope of the known line is the coefficient of 'x' which is .
Therefore, the slope of the unknown line is also
Now, for a line with slope 'm' and a point on it is given as:
Here, . Therefore,
Hence, the equation of the unknown line is .
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)