Answer:
B. 300
Step-by-step explanation:
Final figure volume = big cuboid volume - small cuboid volume
Volume of a cuboid = height x width x length
Final figure = (6 x 6 x 15) - (4 x 4 x 15)
Final figure = 540 - 240
Final figure = 300mm^2
Answer:
62.17% probability that a randomly selected exam will require more than 15 minutes to grade
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:

What is the probability that a randomly selected exam will require more than 15 minutes to grade
This is 1 subtracted by the pvalue of Z when X = 15. So



has a pvalue of 0.3783.
1 - 0.3783 = 0.6217
62.17% probability that a randomly selected exam will require more than 15 minutes to grade
Answer:
no
Step-by-step explanation:
Answer:
<h3>The possibilities of length and width of the rectangle are </h3><h3>x=1, y=0.24;</h3><h3>x=0.5, y=0.48;</h3><h3>x=0.25, y=0.96;</h3><h3>x=2, y=0.12</h3>
Step-by-step explanation:
Given that the area is 0.24 square meter
The area of a rectangle is given by
square units
Let x be the length and y be the width.
Since the area is 0.24 square meter, we have the equation:
, with x and y measures in meters
If we want to know some possibilities of x and y, we can assume a value for one of them, and then calculate the other one using the equation.
Now choosing some values for "x", we have:
Put x = 1

∴ y = 0.24
Now put x = 0.5 we get
∴ y = 0.48
Put x = 0.25
∴ y = 0.96
Put x = 2

∴ y = 0.12
Y = e^tanx - 2
To find at which point it crosses x axis we state that y= 0
e^tanx - 2 = 0
e^tanx = 2
tanx = ln 2
tanx = 0.69314
x = 0.6061
to find slope at that point first we need to find first derivative of funtion y.
y' = (e^tanx)*1/cos^2(x)
now we express x = 0.6061 in y' and we get:
y' = k = 2,9599