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andrezito [222]
3 years ago
5

If each of the numbers in the following data set were multiplied by 18, what would be the new median of the data set? 14, 20, 18

, 58, 71,36, 28
A. 360
B. 504
C. 648
D. 324​
Mathematics
1 answer:
Zigmanuir [339]3 years ago
6 0

Answer:

Step-by-step explanation:

Arange the numbers in ascending order

14,18,20,28,36,58,71

You will see that the median of these data set is 28. So when you multiple it by 18, the answer will be 504.

Therefore, the answer is B. 504

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Solve the problem c=3.14d for d
Maksim231197 [3]

Answer:

c = 3.14d \\ d =  \frac{c}{3.14}

It's change of subject. :)

3 0
2 years ago
What is the greatest ten you can multiply by 2 to get close to but not over 58?
ExtremeBDS [4]

Answer:

20

Step-by-step explanation:

20 x 2 =40

40 is under 58 and the closest to 58

the next 10 is 30 and 30 x 30 is over 58 with 60

3 0
2 years ago
I need help with my math homework
adoni [48]

Answer:

96.52 centimeters

Step-by-step explanation:

1 inch is equal to 2.54 centimeters

38 inches = 38*2.54 cm

38 * 2.54 = 96.52

The answer is 96.52 centimeters

Hope this helps :)

Have a great day

8 0
3 years ago
Read 2 more answers
Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2
IRISSAK [1]

(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0

Suppose the ODE has a solution of the form F(x,y)=C, with total differential

\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0

This ODE is exact if the mixed partial derivatives are equal, i.e.

\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}

We have

\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)

\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)

so the ODE is indeed exact.

Integrating both sides of

\dfrac{\partial F}{\partial x}=(x+y)^2

with respect to x gives

F(x,y)=\dfrac{(x+y)^3}3+g(y)

Differentiating both sides with respect to y gives

\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2

\implies g(y)=-\dfrac{y^3}3-2y+C

\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C

so the general solution to the ODE is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C

Given that y(1)=1, we find

\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13

so that the solution to the IVP is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13

\implies\boxed{(x+y)^3-y^3-6y=1}

5 0
3 years ago
Given f(x) =3x -1 and g(x) = 2x-3, which value of x does g(x)=f(2)
emmainna [20.7K]

f(x) = 3x - 1

g(x) = 2x - 3


g(x) = f(2)

g(x) = 3.2 - 1

g(x) = 6 - 1

g(x) = 5


If g(x) = 5, then, 2x - 3 = 5


2x - 3 = 5

2x = 5 + 3

2x = 8

x = 8/2

x = 4


So,


f(2) = g(4)

4 0
3 years ago
Read 2 more answers
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