Answer:
YES
Step-by-step explanation:
Any straight line can be represented by an equation of the form
Y = mX + b
where m is the slope and b is the Y intercept, so these two equations are lines. The only solution is where the two lines cross, so must satisfy both equations. Plugging in 3 for X and -2 for Y we get
Y = -3X + 7
-2 = -3(3) + 7
-2 = -9 + 7
-2 = -2 True
AND,
Y = 2X - 8
-2 = 2(3) - 8
-2 = 6 - 8
-2 = -2 True
So since both are true, yes (3, -2) is the solution.
Part A: x = -5/4, 3 || (-5/4, 0) (3, 0)
To find the x-intercepts, we need to know where y is equal to 0. So, we will set the function equal to 0 and solve for x.
4x^2 - 7x - 15 = 0
4 x 15 = 60 || -12 x 5 = 60 || -12 + 5 = -7
4x^2 - 12x + 5x - 15 = 0
4x(x - 3) + 5(x - 3) = 0
(4x + 5)(x - 3) = 0
4x + 5 = 0
x = -5/4
x - 3 = 0
x = 3
Part B: minimum, (7/8, -289/16)
The vertex of the graph will be a minimum. This is because the parabola is positive, meaning that it opens to the top.
To find the coordinates of the parabola, we start with the x-coordinate. The x-coordinate can be found using the equation -b/2a.
b = -7
a = 4
x = -(-7) / 2(4) = 7/8
Now that we know the x-value, we can plug it into the function and solve for the y-value.
y = 4(7/8)^2 - 7(7/8) - 15
y = 4(49/64) - 49/8 - 15
y = 196/64 - 392/64 - 960/64
y = -1156/64 = -289/16 = -18 1/16
Part C:
First, start by graphing the vertex. Then, use the x-intercepts and graph those. At this point we should have three points in a sort of triangle shape. If we did it right, each of the x-values will be an equal distance from the vertex. After we have those points graphed, it is time to draw in the parabola. Knowing that the parabola is positive, we draw in a U shape that passes through each of the three points and opens toward the top of the coordinate grid.
Hope this helps!
Answer:
There is a 2% probability that the student is proficient in neither reading nor mathematics.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a student is proficient in reading
B is the probability that a student is proficient in mathematics.
C is the probability that a student is proficient in neither reading nor mathematics.
We have that:
In which a is the probability that a student is proficient in reading but not mathematics and 
is the probability that a student is proficient in both reading and mathematics.
By the same logic, we have that:
Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So
In which
65% were found to be proficient in both reading and mathematics.
This means that 
78% were found to be proficient in mathematics
This means that 
85% of the students were found to be proficient in reading
This means that 
Proficient in at least one:
What is the probability that the student is proficient in neither reading nor mathematics?
There is a 2% probability that the student is proficient in neither reading nor mathematics.
Answer:
e_3^{4} will be the exponential expression.
Step-by-step explanation:
so sorry i can't help explain!!
I'm pretty sure it would be <span>x - y + 1 = 0 because if you plug in one of the coordinates into each equation, this is the only one that works.
</span><span>y - x + 1 = 0
2 - 1 + 1 = 0
2 = 0 not correct
</span><span>x - y + 1 = 0
1 - 2 + 1 = 0
0 = 0 correct
</span><span>-x - y + 1 = 0
-1 - 2 + 1 = 0
-2 = 0 not correct
</span>
