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sineoko [7]
3 years ago
15

A friend has a 78% average before the final exam for a course. That score includes everything but the final, which counts for 20

% of the course grade.
What is the best course grade your friend can earn?
Mathematics
1 answer:
Karolina [17]3 years ago
4 0

Answer:

98%

Step-by-step explanation:

im not real sure but i think thats it if im wrong srry

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What is the area of a rectangle with two sides of length 1 and diagonal length 2?
gtnhenbr [62]
Other two sides = x

Then, x² + 1² = 2²

x² = 4 - 1
x = √3

Now, Area = Length * Width
A = 1 * √3
A = √3

In short, Your Answer would be: √3 or 1.73

Hope this helps!
7 0
3 years ago
If 12 is replaced with 3 in the following set, what will happen to the value of the interquartile range? 28, 45, 12, 34, 36, 45,
crimeas [40]
The answer  would be (it would stay the same)
5 0
3 years ago
Find the area of the shaded region.
luda_lava [24]

Answer:

The area of the shaded region is 11.6\ cm^{2}

Step-by-step explanation:

we know that

The area of the shaded region is equal to the area of the sector of circle of angle 68.9 degrees minus the area of the isosceles triangle

step 1

Find the area of sector of the circle

The area of circle is equal to

A=\pi r^{2}

assume

\pi =3.14

r=9.28\ cm

substitute

A=(3.14)(9.28)^{2}

A=270.41\ cm^{2}

Remember that the area of a circle subtends a central angle of 360 degrees

so

using proportion Find out the area of a sector with a central angle of 68.90 degrees

Let

x -----> the area of a sector

270.41/360=x/68.90\\\\x=68.90*270.41/360\\\\x=51.75\ cm^{2}

step 2

Find the area of the isosceles triangle

Applying the law of sines

The area is equal to

A=(1/2)r^{2}sin(68.90)

we have

r=9.28\ cm

substitute

A=(1/2)(9.28)^{2}sin(68.90)=40.17\ cm^{2}

step 3

Find the area of the shaded region

51.75-40.17=11.58\ cm^{2}

Round to the nearest tenth

11.58=11.6\ cm^{2}

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3 years ago
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Deffense [45]
3.3m
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4 0
3 years ago
What is 120 as a decimal
Agata [3.3K]
120.0 hope his was helpful
5 0
3 years ago
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