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Shtirlitz [24]
3 years ago
7

Cathrine was concerned about her report card grade. She had 5 tests the whole semester. If her scores on each test were 93, 82,

74, 92 and 67 what is the mean/average of her grade?
Mathematics
1 answer:
enyata [817]3 years ago
4 0

Answer:

81.6

Step-by-step explanation:

mean/average of her grade = sum of all scores/the number of scores

= (93+82+74+92+67)/5

= 81.6

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No one helped so im asking for these 2 again
ipn [44]

Answer:

3. Option No. 2:  

6x + 2y = 22

3x + 3y = 18

4. Option No. 2:

$2.50

Step by step explanation:

4. 6x + 2y = 22  

 (3x + 3y = 18) * 2

- Negate

6x + 2y = 22

6x + 6y = 36

2y - 6y = 22 - 36

-4y = -14

y = -14/-4

y = 3.5

An order of onion rings = $3.50

- Substitute to find for x

6x + 2(3.5) = 22

6x + 7 = 22

6x = 22 - 7

6x = 15

x = 15/6

x = 2.5

An order of French fries = $2.50

3 0
3 years ago
in this triangle, what is the value of x? Enter your answer, rounded to the nearest tenth, in the box. x = A right triangle with
Marat540 [252]

Answer:

56.7°

Step-by-step explanation:

We solve this above question using the Trigonometric function of Sine

sin x = Opposite side/Hypotenuse

Opposite = 66mm

Hypotenuse = 79 mm

sin x = 66/79

x =arc sin(66/79)

x = arcsin(0.8354430379746836)

x = 56.661998553°

Approximately = 56.7°

8 0
3 years ago
Read 2 more answers
Circle O, with center (x, y), passes through the points A(0, 0), B(–3, 0), and C(1, 2). Find the coordinates of the center of th
balandron [24]
The answer:

the main formula of the circle's equation is 
(x-a)²+ (y-b)² = R²
where C(a, b) is the center of the circle
R is the radius

if a point A(x', y') passes through the circle, so the equation of the circle can be written as 
(x'-a)²+ (y'-b)² = R², and that is  a main formula.


<span>Circle O, with center (x, y), passes through the points A(0, 0), B(–3, 0), and C(1, 2), so we have exactly three equation:
</span>
(0-x)² + (0-y)² = R², circle O passes through A
x²+y²= R²
(-3 -x)² + (0-y)² = R², circle O passes through B
(-3 -x)² + (y)² = R²
(1-x)² + (2-y)² = R², circle O passes through A
(1-x)² + (2-y)² = R²

and we know that R= OA = OC= OB, 
OA=R= sqrt( (0-x)² + (0-y)² ) = OB = sqrt((-3 -x)² + (0-y)²), this implies

x²+y² = (-3 -x)² + (0-y)² , it implies x² = 9+ x² + 6x , and then -9/6=x, x= -3/2
and when OA = OC
x²+y² =(1-x)² + (2-y)²  so, x²+y² =1+x²-2x +4+y²-4y, therefore -5= -2x -4y
 -5= -2x -4y, when x = -3 /2   we obtain y = 2

the center is C(-3/2, 2)
7 0
4 years ago
QUESTION 11.1
lawyer [7]

The group paid $ 5250 at first city and $ 6250 at second city

<u>Solution:</u>

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the  second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the  tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

<em><u>Therefore, a equation is framed as:</u></em>

8.5 % of x + 5.5 % of y = 790

\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790

0.085x + 0.055y = 790 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

<h3>x = 5250</h3>

<em><u>Substitute x = 5250 in eqn 1</u></em>

y = 5250 + 1000

<h3>y = 6250</h3>

Thus the group paid $ 5250 at first city and $ 6250 at second city

8 0
3 years ago
God bless everyone (ㆁωㆁ)
beks73 [17]
૮₍˶ᵔ ᵕ ᵔ˶₎ა
God bless you too
7 0
3 years ago
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