I need help with this calculus question
2 answers:
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For the cylinder on the left-hand-side,
now, for the triangular prism on the right-hand-side,
notice is really just 2 triangles and 3 rectangles, stacked up to each other at the edges.
the triangles have a base of 4, and a height of 1.5.
the rectangle on the left and the one on the right is a 6x2.5 rectangle.
the rectangle at the bottom, is a 4x6 rectangle.
adding their areas, is the area of the prism,
![\bf \stackrel{two~triangles}{2\left[\cfrac{1}{2}(4)(1.5) \right]}+\stackrel{left-right~rectangles}{2(6\cdot 2.5)}+\stackrel{bottom~rectangle}{4\cdot 6} \\\\\\ 6~~+~~30~~+~~24](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Btwo~triangles%7D%7B2%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%284%29%281.5%29%20%20%5Cright%5D%7D%2B%5Cstackrel%7Bleft-right~rectangles%7D%7B2%286%5Ccdot%202.5%29%7D%2B%5Cstackrel%7Bbottom~rectangle%7D%7B4%5Ccdot%206%7D%0A%5C%5C%5C%5C%5C%5C%0A6~~%2B~~30~~%2B~~24)
now, for the triangular prism on the right-hand-side,
notice is really just 2 triangles and 3 rectangles, stacked up to each other at the edges.
the triangles have a base of 4, and a height of 1.5.
the rectangle on the left and the one on the right is a 6x2.5 rectangle.
the rectangle at the bottom, is a 4x6 rectangle.
adding their areas, is the area of the prism,
Answer:
And for this case we know this condition:
By the complement rule we know that:
But since the distribution is symmetrical we know that:
So then the statement for this case is FALSE.
b. False
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
For this case if we define the random variable of interest X and we know that this random variable follows a normal distribution:
And for this case we know this condition:
By the complement rule we know that:
But since the distribution is symmetrical we know that:
So then the statement for this case is FALSE.
b. False