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Mumz [18]
3 years ago
10

I have a 74% in my class and I need a 78% to pass what do I need to get on my final exam?

Mathematics
1 answer:
Mkey [24]3 years ago
3 0
So the 74% that you have right now is the other 80% of your grade. So you multiply it by its weight and solve for the other one making it equal to 78%.

(.74)(.8) + (.2)(x) = .78
 .592 + (.2)(x) = .78
    (.2)(x) = .188
       (x) = .94

So, you need to get a 94% on your final test.

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The area of the yellow region is 112 cm^2. find the value of x​
mrs_skeptik [129]
<h3>Answer:   x = 7</h3>

===========================================================

Explanation:

The largest rectangle (composed of the green and yellow sections combined) has area of 11*12 = 132 cm^2.

The yellow region takes up 112 of those 132 sq cm. This must mean the green region takes up 132-112 = 20 cm^2.

The horizontal portion of the green rectangle is 12-x cm. The vertical portion is 11-x cm. We can form the area of the green rectangle as an algebraic expression like so

area = length*width

area = (11-x)*(12-x)

area = 132 - 11x - 12x + x^2 .... apply the FOIL rule

area = x^2 - 23x + 132

Set this equal to the 20 cm^2 we found earlier.

x^2 - 23x + 132 = 20

x^2 - 23x + 132-20 = 0

x^2 - 23x + 112 = 0

We could factor or we could use the quadratic formula. I'll go with the second option.

We'll plug in a = 1, b =  -23, c = 112

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-23)\pm\sqrt{(-23)^2-4(1)(112)}}{2(1)}\\\\x = \frac{23\pm\sqrt{81}}{2}\\\\x = \frac{23\pm9}{2}\\\\x = \frac{23+9}{2}\ \text{ or } \ x = \frac{23-9}{2}\\\\x = \frac{32}{2}\ \text{ or } \ x = \frac{14}{2}\\\\x = 16\ \text{ or } \ x = 7\\\\

One of these solutions isn't feasible. Note how if x = 16, then this exceeds both the 11 cm and 12 cm sides. So this x value is not possible.

However, x = 7 is possible.

If x = 7, then the horizontal portion of the green rectangle is 12-x = 12-7 = 5 cm. Also, the vertical portion of the green rectangle would be 11-x = 11-7 = 4 cm. The area then is length*width = 5*4 = 20 cm^2 which matches up with what we got earlier. So the answer is confirmed.

7 0
3 years ago
8/11(n-10)=64 I need to show my work to
Bogdan [553]
Hope you understand it and see what I did.

6 0
3 years ago
Read 2 more answers
Wire of length 20m is divided into two pieces and the pieces are bent into a square and a circle. How should this be done in ord
blondinia [14]

Answer:

The radius of the circle ,r= 1.43 m

The length of the side of square ,a= 2.77 m

Step-by-step explanation:

Given that

L= 20  m

Lets take radius of the circle =r m

The total parameter of the circle = 2π r

Area of circle ,A=π r²

The side of the square = a m

The total parameter of the square = 4 a

Area of square ,A'=a²

The total length ,L= 2π r+ 4 a

20  = 2π r+ 4 a

r=3.18 - 0.63 a

The total area = A+ A'

                   A"   =π r² +a²

A"= 3.14(3.18 - 0.63 a)² + a²

For minimize the area

\dfrac{dA"}{da}=0

3.14 x 2(3.18 - 0.63 a) (-0.63) + 2 a = 0

3.14 x (3.18 - 0.63 a) (-0.63) + a = 0

-6.21 + 1.24 a + a=0

2.24 a = 6.21

a=2.77 m

r= 3.18 - 0.63 a

r= 3.18 - 0.63 x 2.77

r=1.43 m

Therefore the radius of the circle ,r= 1.43 m

The length of the side of square ,a= 2.77 m

4 0
3 years ago
Read 2 more answers
A set of N = 20 exam scores has a mean of m = 50. The instructor discovered that an error was made grading one exam, so 20 point
tatuchka [14]

Answer:

The new mean for the exam scores is 51

Step-by-step explanation:

the mean can be calculated using the formula

mean of scores = \frac{Sum total    of scores}{Number of students}

before the 20 points was added, the mean of the score was = 50.

Inserting this into the formula, we have:

50 = \frac{Sum total}{20 students}

from this, we can compute the sum total of the scores as

sum total = 50 × 20 = 1000

when the extra 20 marks is added, the sum of the entire scores will be changed to 1000 + 20 = 1020.

Hence the new mean will be computed as (New Sum total of scores / Number of students)

= \frac{1020}{20} = 51.

∴ The new mean of the scores is = 51

8 0
3 years ago
Please help me with these two problems thank you!!​
andriy [413]

Answer:

1st question: y=1/3x-1

2nd question: y=-2/3x+9

Step-by-step explanation:

Y=mx+b form is what you are using

when a line in parallel to a line they have the same slop wich is m

when a line in perpendicular they have the opposite reciprocal

Examples: 2/5 opposite reciprocal would be -5/2

8 0
3 years ago
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