Answer:
10.20% probability that a randomly chosen book is more than 20.2 mm thick
Step-by-step explanation:
Problems of normally distributed samples are 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:
250 sheets, each sheet has mean 0.08 mm and standard deviation 0.01 mm.
So for the book.
What is the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers)
This is 1 subtracted by the pvalue of Z when X = 20.2. So
has a pvalue of 0.8980
1 - 0.8980 = 0.1020
10.20% probability that a randomly chosen book is more than 20.2 mm thick
Answer:
So I can't actually do this at the moment but I'll try to explain what to do
first find the area of the rectangle (12x16)
then, the confusing part, pretend the semicircle is a full circle and find the area and divide by 2
add the two up and put 
and that should be it
Ok I can help just tell me the problem
Answer:
Hyperbola
Step-by-step explanation:
The polar equation of a conic section with directrix ± d has the standard form:
r=ed/(1 ± ecosθ)
where e = the eccentricity.
The eccentricity determines the type of conic section:
e = 0 ⇒ circle
0 < e < 1 ⇒ ellipse
e = 1 ⇒ parabola
e > 1 ⇒ hyperbola
Step 1. <em>Convert the equation to standard form
</em>
r = 4/(2 – 4 cosθ)
Divide numerator and denominator by 2
r = 2/(1 - 2cosθ)
Step 2. <em>Identify the conic
</em>
e = 2, so the conic is a hyperbola.
The polar plot of the function (below) confirms that the conic is a hyperbola.
Answer:
Yeah, theres no way
Step-by-step explanation:
lol
