Is the ordered pair (6,5) a solution to the equation 3x

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Nataly [62]

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4 0

3 years ago

Max adds 7 to a number the. Múltiples the sum by -4 the result is 3 times the same number write and since an equation to find th

GrogVix [38]

Answer:

The number is -4.

Step-by-step explanation:

Let the number be n.

Given:

Max adds 7 to a number.

hence it is given as n+7

Multiply's the sum by -4 the result is 3 times the same number.

Hence the equation can be written as;

$-4(n+7)=3n$

Now Solving the above equation to find value of n we get;

$-4(n+7)=3n\\-4n-28=3n\\-4n-3n=28\\-7n=28\\n=\frac{28}{7}=-4$

The value of n is -4.

Now when we add 7 to number -4 we get answer as 3.

And when the sum is multiplied by -4 we get answer -12.

Also 3 times of number is equal to 3 multiplied by -4 we get answer as -12.

Hence when the sum is multiplied by -4 it is equal to 3 times of same number.

Hence from above we can say that the number is -4.

7 0

3 years ago

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 12.1 fluid ounc

Ludmilka [50]

Answer:

$\mu = 12.1\\\sigma = 0.3$

a) Find the probability that the drink is less than 11.9 fluid ounces

We are supposed to find P(x<11.9)

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

Refer the z table

So, P(z<-0.67) = 0.2514

Hence the probability that the drink is less than 11.9 fluid ounces is 0.2514

b)Find the probability that the drink is between 11.6 and 11.9 fluid ounces

x= 11.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.6-12.1}{0.3}$

$z=-1.67$

x= 11.9

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{11.9-12.1}{0.3}$

$z=-0.67$

We are supposed to find P(11.6<x<11.9)

So, P(11.6<x<11.9)= P(x<11.9)-P(x<11.6)

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

Refer the z table

P(-1.67<z<-0.67)= P(z<-0.67)-P(z<-1.67)

P(-1.67<z<-0.67)= 0.2514-0.0475

P(-1.67<z<-0.67)= 0.2039

So, the probability that the drink is between 11.6 and 11.9 fluid ounces is 0.2039.

c)Find the probability that the drink is more than 12.6 fluid ounces

We are supposed to find P(x>12.6)

x= 12.6

We will use z score formula :

$z=\frac{x-\mu}{\sigma}$

$z=\frac{12.6-12.1}{0.3}$

$z=1.67$

Refer the z table

P(x>12.6)=1-P(x<12.6)

P(z>1.67)=1-P(z<1.67)

P(z>1.67)=1-0.9525

P(z>1.67)=0.0475

Hence the probability that the drink is more than 12.6 fluid ounces is 0.0475

d) Is the drink containing more than 12.6 fluid ounces an unusual event?

Since p value is less than alpha (0.05)

Answer = No, because the probability that a drink containing more than 12.6 fluid ounces is less than 0.05, this event is not unusual.

3 0

3 years ago

Solve the following differential equation: (1− 5 y +x) dy/dx +y= 5/x −1 .<br><br> C=

Ganezh [65]

Answer:

$y(1+x)+x-5\ln xy=C$

Step-by-step explanation:

Consider the given differential equation is

$(1-\frac{5}{y}+x)\frac{dy}{dx}+y=\frac{5}{x}-1$

$(1-\frac{5}{y}+x)\frac{dy}{dx}=\frac{5}{x}-1-y$

$(1-\frac{5}{y}+x)dy=(\frac{5}{x}-1-y)dx$

Taking all variables on right sides.

$(1-\frac{5}{y}+x)dy-(\frac{5}{x}-1-y)dx=0$

$(-\frac{5}{x}+1+y)dx+(1-\frac{5}{y}+x)dy=0$

Let as assume,

$M=-\frac{5}{x}+1+y$ and $N=1-\frac{5}{y}+x$

Find partial derivatives $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$

$\frac{\partial M}{\partial y}=1$ and $\frac{\partial N}{\partial x}=1$

Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ , therefore the given differential equation is exact.

The solution of the exact differential equation is

$\int Mdx+\int N(\text{without x)}dy=C$

$\int (-\frac{5}{x}+1+y)dx+\int (1-\frac{5}{y})dy=C$

$yx-5\ln x+x+y-5\ln y=C$

$y+x+xy-5\ln x-5\ln y=C$

$y(1+x)+x-5(\ln x+\ln y)=C$

$y(1+x)+x-5\ln xy=C$

7 0

3 years ago

Use substitution to solve the following system of equations.<br> y=3x +8<br> 5x+ 2y = 5

Shkiper50 [21]

Answer:

5x+2(3x+8)

5x+6x+16

11x+16

11x=-16

X=16/11

6 0

3 years ago

Is the ordered pair (6,5) a solution to the equation 3x - 4y = -2? Explain.