Tion 15/30

1 answer:

6 0

The percentage of students that passed the exam is 25.2%

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<u>The given parameters;</u>

The number of the male students that pass the exam;

$= 0.2 \times 85 \\\\= 17$

The number of the female students that pass the exam;

$= 0.4 \times 30 \\\\= 12$

The total number of students in the school;

= 85 + 30

= 115

The total number of the students that passed the exam;

= 17 + 12

= 29

The percentage of students that passed the exam;

$= \frac{total \ number \ that \ passed }{total \ number \ of \ the \ students} \times 100\% \\\\= \frac{29}{115} \times 100\%\\\\= 25.2 \%$

Thus, the percentage of students that passed the exam is 25.2%

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Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .