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Dafna1 [17]
3 years ago
9

In a survey three out of seven people named blue as their favorite color. One out of six named red. If 1092 people were included

in the survey, how many named neither blue nor red as their favorite color?
Mathematics
2 answers:
Marysya12 [62]3 years ago
7 0

Answer:

442 I belive

Step-by-step explanation:

MrMuchimi3 years ago
3 0

Answer:

<h2>442 people named neither blue nor red.</h2>

Step-by-step explanation:

Givens:

  • \frac{3}{7} people named blue.
  • \frac{1}{6} people named red.
  • The total number of people is 1092

So, we first have to calculate the number of people that named blue and red.

\frac{3}{7}1092=468\\\frac{1}{6}1092=182

This means that 468+182=650 people named blue or red.

If we find the difference between 650 and 1092, we would have the total number of people that didn't named blue nor red.

1092-650=442.

Therefore, 442 people named neither blue nor red.

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Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.
lakkis [162]

Answer:

2*sin(x)+y*cos(x)-cos(y)=C_1

Step-by-step explanation:

Let:

P(x,y)=2*cos(x)-y*sin(x)

Q(x,y)=cos(x)+sin(y)

This is an exact differential equation because:

\frac{\partial P(x,y)}{\partial y} =-sin(x)

\frac{\partial Q(x,y)}{\partial x}=-sin(x)

With this in mind let's define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x}=P(x,y)

and

\frac{\partial f(x,y)}{\partial y}=Q(x,y)

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y)

f(x,y)=\int\  2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}

Now, let's replace the previous result into \frac{\partial f(x,y)}{\partial y}=Q(x,y) :

cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)

Solving for \frac{dg(y)}{dy}

\frac{dg(y)}{dy}=sin(y)

Integrating both sides with respect to y:

g(y)=\int\ sin(y)  \, dy =-cos(y)

Replacing this result into f(x,y)

f(x,y)=2*sin(x)+y*cos(x)-cos(y)

Finally the solution is f(x,y)=C1 :

2*sin(x)+y*cos(x)-cos(y)=C_1

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3 years ago
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