Jean-Paul incorrectly states that
1 answer:
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Let
x-------> the length of the rectangle
y------> the width of the rectangle
we know that
The area of the rectangle is equal to
The area of the two congruent right triangles is equal to the area of the rectangle
so
-------> equation A
-----> equation B
Substitute equation B in equation A
--------> equation that can be used to solve for the length of the rectangle
Using a graph tool-------> solve the quadratic equation
see the attached figure
The solution is
-----> the length of the rectangle
Find the value of y
----> the width of the rectangle
Statements
<u>case A)</u> The area of the rectangle is square inches
The statement is True
See the procedure
<u>Case B)</u> The equation can be used to solve for the dimensions of the triangle
The statement is False
Because, the equation can be used to solve for the dimensions of the triangle
<u>case C)</u> The equation can be used to solve for the length of the rectangle
The statement is True
see the procedure
<u>case D)</u>The triangle has a base of inches and a height of
inches
The statement is True
Because, the base of the triangle is equal to the length of the rectangle and the height of the triangle is equal to the width of the rectangle
<u>case E)</u> The rectangle has a width of inches
The statement is False
See the procedure
Answer: 20
Step-by-step explanation:
We assume that the heights of boys in a high school basketball tournament are normally distributed.
Given : Mean height of boys : inches.
Standard deviation: inches.
Let x denotes the heights of boys in a high school basketball tournament .
Then the probability that a boy is taller than 70 inches will be :-
Now, the expected number of boys in a group of 40 who are taller than 70 inches will be :-
Hence, the expected number of boys in a group of 40 who are taller than 70 inches=20
Answer:
Step-by-step explanation:
the equation of said line would be y=2x +3
it would look like this graph
