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Maru [420]
3 years ago
6

Cna i get some help with dis plz

Mathematics
1 answer:
Tomtit [17]3 years ago
6 0

Let b_1,b_2,\ldots,b_{20} be the 20 marks of the boys, and g_1,g_2,\ldots,g_{10} be the 10 marks of the girls.

We know that the global mean was 70, meaning that

\dfrac{b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}}{30}=70

Multiplying both sides by 30 we deduce that the sum of the scores of the whole classroom is

b_1+b_2+\ldots+b_{20}+g_1+g_2+\ldots+g_{10}=2100

By the same logic, we work with the marks of the boys alone: we know the average:

\dfrac{b_1+b_2+\ldots+b_{20}}{20}=62

And we deduce the sum of the marks for the boys:

b_1+b_2+\ldots+b_{20}=1240

Which implies that the sum of the marks of the girls is 2100-1240=860

And finally, the mean for the girls alone is

\dfrac{860}{10}=86

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blondinia [14]

f(x)=mx+b\to y=mx+b

We have two points:

(-3,\ 0);\ (-4,\ 7)

The point slope form of the line:

y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}

Substitute

m=\dfrac{7-0}{-4-(-3)}=\dfrac{7}{-1}=-7

y-0=-7(x-(-3))\\\\y=-7(x+3)\\\\y=-7x-21

Answer:

f(x)=-7x-21

8 0
3 years ago
25, 8, 10, 35, 5, 45, 40, 30, 20 What is the upper quartile of the data set above?A.35
Vlada [557]
<span>An upper quartile is the range of numbers above the median in a set. Thus, the numbers have to be rearranged mentally or on paper. Fortunately, there are an odd set of numbers, so one number, 25, is the median. The upper quartile is 30, 35, 40, 45.</span>
6 0
3 years ago
2.199 rounded to the nearest whole number
steposvetlana [31]
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3 0
3 years ago
Recall the equation for a circle with center ( h , k ) (h,k) and radius r r. At what point in the first quadrant does the line w
garri49 [273]
Equation of this circle is
x^2 + (y - 2)^2 = 36
y = 2.5x + 2

Substitute for y in the equation of the circle:-

x^2 + (2.5x + 2 - 2)^2 = 36

x^2 + 6.25x^2 = 36

x^2 = 36 / 7.25  

x = +/-   6  /  2.693  =  +/- 2.228

when x = 2.228 y = 2.5(2.228) + 2 =  7.57    to nearest hundredth

when x = -2.228 y = 2.5(-2.228) + 2 =   -3.57

So they intersect at 2 points but the intersect in the first quadrant is at (2.23, 7,57)        to nearest hundredth.
7 0
4 years ago
18c) Find the value of x in each case.
iVinArrow [24]

Answer:

x = 22.5°

Step-by-step explanation:

m∠DEA = 2x

DE || AB

m∠EAB = 2x  - by Alternate Interior Angles

Exterior angle = 6x, which equals 2x + 4x

m∠BEA = 4x

Exterior Angle 4x = 2x + 2x

m∠B = 2x

2x + 4x + 2x = 180°

8x = 180°

x = 22.5°

<h2>Exterior Angles of a Triangle:</h2>

An exterior angle of a triangle is equal to the two REMOTE angles inside the triangle.

In this example, we know that 6x has remote angles of ∠EAB and ∠BEA

We found out that m∠EAB = 2x, so m∠BEA must equal 6x - 2x or 4x

<h2>Sum of angles in a Triangle:</h2>

The sum of all angles in a triangle is equal to 180°.

In this example, we found all the angles, 2x, 4x, 2x. We added all of those values to equal 8x

Since the sum of all angles = 180°:

8x = 180°

x = 180°/8

x = 22.5°

-Chetan K

8 0
3 years ago
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