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dalvyx [7]
3 years ago
11

5ry - c = q solve for r

Mathematics
1 answer:
Tems11 [23]3 years ago
4 0

Answer:

r = (q + c)/5y

Step-by-step explanation:

in order to solve this expression thereby expressing r in terms of y,c and q.

5ry - c = q

from the above expression

5ry = q + c

divide both sides by 5y

5ry/5y = q + c/5y

r = (q + c)/5y

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7 0
4 years ago
Read 2 more answers
1. A researcher would like to learn about the difference in proportions of students and faculty who support textbook price caps.
Colt1911 [192]

Answer: (0.186, 0.298)

Step-by-step explanation:

The formula to find the confidence interval for the difference of the population proportion is given by :-

\hat{p}_1-\hat{p}_2\pm z^*\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}   ... (i)

, where n_1= Sample size of population 1.

n_2=  Sample size of population 2.

\hat{p_1}= Sample proportion of population 1.

\hat{p}_2 = Sample proportion of population 2.

z* = Critical z-value corresponding to confidence interval

As per given , we have

n_1=n_2=500

\hat{p}_1=\dfrac{378}{500}=0.756\\\hat{p}_2=\dfrac{256}{500}=0.512

Critical value corresponds to 95% confidence interval = 1.96

Put all these values , in (i) , we get

0.756-0.512\pm 1.96\sqrt{\dfrac{0.756(1-0.756)}{500}+\dfrac{0.512(1-0.512)}{500}}\\\\=0.244\pm1.96(\sqrt{0.00086864})\\\\=0.244\pm1.96(0.0294727)\\\\=0.244\pm0.0577665\\\\=(0.244-0.0577664,\ 0.24+0.0577664)\\\\=(0.1862336,\ 0.2977664)\approx(0.186,\ 0.298)

Hence, the 95% confidence interval for the difference of the population proportions= (0.186, 0.298)

6 0
3 years ago
If 8 out of 20 students in a class are boys,what percent of the class is made up of boys
lara31 [8.8K]

Answer:

40%

Step-by-step explanation:

First, divide 100 by 20. You do this because you need something out of 100 to equal a percent. (Example- 50 out of 100 is 50%)

Next you need to take the number you got from dividing (5) and multiply that with 8. This leaves you with 40/100 or 40%.

Hope this helps!

-Coconut;)

3 0
3 years ago
Please anyone help me ASAP
OLga [1]
A nice, interesting question. We have to be known to a equation called as the Circle equation. It is given by the formula of:

\boxed{\mathbf{(x - a)^2 + (y - b)^2 = r^2}}

That is the circle equation with a representation of the variable "a" and variable "b" as the points for the circle's center and the variable of "r" is representing the radius of the circle.

We are told to convert the given equation expression into a typical standard format of circle equation. This will mean we can easily deduce the values of the following variables and/or the points of the circle including the radius of the circle by our standard circle equation via conversion of this expression. So, let us start by interpreting this through equation editor for mathematical expression LaTeX, for a clearer view and better understanding.

\boxed{\mathbf{Given \: \: Equation: x^2 + y^2 - 4x + 6y + 9 = 0}}

Firstly, shifting the real numbered values or the loose number, in this case it is "9", to the right hand side, since we want an actual numerical value and the radius of circle without complicating and stressing much by using quadratic equations. So:

\mathbf{x^2 - 4x + 6y + y^2 = - 9}

Group up the variables of "x" and "y" for easier simplification.

\mathbf{\Big(x^2 + 4x \Big) + \Big(y^2 + 6y \Big) = - 9}

Here comes the catch of applying logical re-squaring of variables. We have to convert the variable of "x" into a "form of square". We can do this by adding up some value on the grouped variables as separately for "x" and "y" respectively. And add the value of "4" on the right hand side as per the square conversion. So:

\mathbf{\Big(x^2 - 4x + 4 \Big) + \Big(y^2 + 6y \Big) = - 9 + 4}

We can see that; our grouped variable of "x" is exhibiting the square of expression as "(x - 2)^2" which gives up the same expression when we square "(x - 2)^2". Put this square form back into our current Expressional Equation.

\mathbf{(x - 2)^2 + \Big(y^2 + 6y \Big) = - 9 + 4}

Similarly, convert the grouped expression for the variable "y" into a square form by adding the value "9" to grouped expression of variable "y" and adding the same value on the right hand side of the Current Equation, as per the square conversion.

\mathbf{(x - 2)^2 + \Big(y^2 + 6y + 9 \Big) = - 9 + 4 + 9}

Again; We can see that; our grouped variable of "y" is exhibiting the square of expression as "(y + 3)^2" which gives up the same expression when we square "(y + 3)^2". Put this square form back into our current Expressional Equation.

\mathbf{(x - 2)^2 + (y + 3)^2 = - 9 + 13}

\mathbf{(x - 2)^2 + (y + 3)^2 = 4}

Re-configure this current Expressional Equational Variable form into the current standard format of Circle Equation. Here, "(y - b)^2" is to be shown and our currently obtained Equation does not exhibit that. So, we do just one last thing. We distribute the parentheses and apply the basics of plus and minus rules. That is, "- (- 3)" is same as "+ (3)". And "4" as per our Circle Equation can be re-written as a exponential form of "2^2"

\mathbf{(x - 2)^2 + \big(y - (- 3) \big)^2 = 2^2}

Compare this to our original standard form of Circle Equation. Here, the center points "a" and "b" are "2" and "- 3". The radius is on the right hand side, that is, "2".

\boxed{\mathbf{\underline{\therefore \quad Center \: \: (a, \: b) = (2, \: - 3); \: Radius \: \: r = 2}}}

Hope it helps.
4 0
3 years ago
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