<span>"Chisel a whittle off the top, please."</span>
Answer:
The cutoff numbers for the number of chocolate chips in acceptable cookies are 16.242 and 27.758
Step-by-step explanation:
Problems of normally distributed samples 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.
In this problem, we have that:
Middle 90%
Between the 50 - (90/2) = 5th percentile to the 50 + (90/2) = 95th percentile.
5th percentile:
X when Z has a pvalue of 0.05. So X when Z = -1.645.
95th percentile:
X when Z has a pvalue of 0.95. So X when Z = 1.645.
The cutoff numbers for the number of chocolate chips in acceptable cookies are 16.242 and 27.758
By definition and properties of the <em>absolute</em> value used on the <em>quadratic</em> equation we conclude that F(|- 4|) = 12.
<h3>How to evaluate a quadratic equation with an absolute value</h3>
Herein we must apply the definition of <em>absolute</em> value prior to evaluating the quadratic equation defined in the statement. From algebra we know that absolute values are defined as:
|x| = x, when x ≥ 0 or - x, when x < 0. (1)
Then, we apply (1) on the quadratic equation:
F(|x|) = |x|² - 2 · |x| + 4
As x < 0, by <em>absolute value</em> properties:
F(|x|) = x² + 2 · x + 4
F(|- 4|) = (- 4)² + 2 · (- 4) + 4
F(|- 4|) = 16 - 8 + 4
F(|- 4|) = 12
By definition and properties of the <em>absolute</em> value used on the <em>quadratic</em> equation we conclude that F(|- 4|) = 12.
To learn more on absolute values: brainly.com/question/1301718
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Answer:
Step-by-step explanation:
One way to determine if the equation has any real solutions is to look at its discriminant. For the equation ax²+bx+c = 0, the discriminant is ...
d = b² -4ac
When the discriminant is negative, both solutions to the quadratic are complex. There are no real solutions in that case.
We can find the discriminant values to be ...
A: d = 2² -4(1)(4) = -12 . . . . no real zeros
B: d = 0² -4(3)(-5) = 60 . . . two real zeros
C: d = 8² -4(-2)(0) = 64 . . . two real zeros
D: d = 10² -4(1)(26) = -4 . . .no real zeros
Expressions A and D have no real zeros.
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<em>Comment on the question</em>
You are given an <em>expression</em>, not an <em>equation</em>. There is no equal sign. Hence, we cannot talk about <em>solutions</em>. We can only talk about <em>zeros</em>, values of x that make the expression have a value of zero.
