All categories

3 years ago

A chorus has 5 girls and 15 boys. If two are chosen at random to sing a duet, what is the probability that both will be boys?

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

Answer:

Step-by-step explanation:

If you ad them together you get 20 15/20 are boys 15 divided by 20 is .75

Galina-37 [17]3 years ago

8 0

.75 or 75%

If there’s 20 kids total in the class and 15 out of the 20 are boys then you would divide the 15 by 20 and get .75 if you were doing it for girls then you would divide 5 by 20 and you would get .25 or 25%

You might be interested in

<img src="https://tex.z-dn.net/?f=3x%20%2B%202%20%3C%20b" id="TexFormula1" title="3x + 2 < b" alt="3x + 2 < b" align="absm

kramer

I could fix it for you.
Merry Christmas

8 0

4 years ago

Factor 3x + 9<br><br> Please explain how you got the answer

eimsori [14]

To factor 3x + 9 you should know that 3 is a common factor of both 3 and 9, so you can factor 3 out in both numbers like this.

3x + 9

3(x + 3)

When you factor out 3 on both numbers you should know that 3 X 3 = 9 and since you factored out 3 there would be another 3 left so you would have the answer: 3(x + 3)

Here is another equation that might help:

24x + 36

Since 12 is a common factor of both numbers, factor 12 out of the equation:

12 (2x + 3)

You know that 12 x 2 = 24 and that 12 x 3 = 36

And since you can't factor out anymore numbers because 2 and 3 doesn't have any common factors with each other you get the answer:

12 (2x + 3)

3 0

3 years ago

Please I need help I been stuck on this problem for 40 minutes

nignag [31]

First
224-108=116miles left
And 4 days left
116/4
=29 miles each day

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Cunderline%7B%20%5Cunderline%7B%20%5Ctext%7Bquestion%7D%7D%7D%20%3A%20" id="TexFormula1"

Inga [223]

Answer:

$y=-\sqrt{3}x+2$

Step-by-step explanation:

We want to find the equation of a straight line that cuts off an intercept of 2 from the y-axis, and whose perpendicular distance from the origin is 1.

We will let Point M be (x, y). As we know, Point R will be (0, 2) and Point O (the origin) will be (0, 0).

First, we can use the distance formula to determine values for M. The distance formula is given by:

$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Since we know that the distance between O and M is 1, d=1.

And we will let M(x, y) be (x₂, y₂) and O(0, 0) be (x₁, y₁). So:

$\displaystyle 1=\sqrt{(x-0)^2+(y-0)^2}$

Simplify:

$1=\sqrt{x^2+y^2}$

We can solve for y. Square both sides:

$1=x^2+y^2$

Rearranging gives:

$y^2=1-x^2$

Take the square root of both sides. Since M is in the first quadrant, we only need to worry about the positive case. Therefore:

$y=\sqrt{1-x^2}$

So, Point M is now given by (we substitute the above equation for y):

$M(x,\sqrt{1-x^2})$

We know that Segment OM is perpendicular to Line RM.

Therefore, their <em>slopes will be negative reciprocals</em> of each other.

So, let’s find the slope of each segment/line. We will use the slope formula given by:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Segment OM:

For OM, we have two points: O(0, 0) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{OM}=\frac{\sqrt{1-x^2}-0}{x-0}=\frac{\sqrt{1-x^2}}{x}$

Line RM:

For RM, we have the two points R(0, 2) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{RM}=\frac{\sqrt{1-x^2}-2}{x-0}=\frac{\sqrt{1-x^2}-2}{x}$

Since their slopes are negative reciprocals of each other, this means that:

$m_{OM}=-(m_{RM})^{-1}$

Substitute:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=-\Big(\frac{\sqrt{1-x^2}-2}{x}\Big)^{-1}$

Now, we can solve for x. Simplify:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=\frac{x}{2-\sqrt{1-x^2}}$

Cross-multiply:

$x(x)=\sqrt{1-x^2}(2-\sqrt{1-x^2})$

Distribute:

$x^2=2\sqrt{1-x^2}-(\sqrt{1-x^2})^2$

Simplify:

$x^2=2\sqrt{1-x^2}-(1-x^2)$

Distribute:

$x^2=2\sqrt{1-x^2}-1+x^2$

So:

$0=2\sqrt{1-x^2}-1$

Adding 1 and then dividing by 2 yields:

$\displaystyle \frac{1}{2}=\sqrt{1-x^2}$

Then:

$\displaystyle \frac{1}{4}=1-x^2$

Therefore, the value of x is:

$\displaystyle \begin{aligned}\frac{1}{4}-1&=-x^2\\-\frac{3}{4}&=-x^2\\ \frac{3}{4}&=x^2\\ \frac{\sqrt{3}}{2}&=x\end{aligned}$

Then, Point M will be:

$\begin{aligned} \displaystyle M(x,\sqrt{1-x^2})&=M(\frac{\sqrt{3}}{2}, \sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2)}\\M&=(\frac{\sqrt3}{2},\frac{1}{2})\end{aligned}$

Therefore, the slope of Line RM will be:

$\displaystyle \begin{aligned}m_{RM}&=\frac{\frac{1}{2}-2}{\frac{\sqrt{3}}{2}-0} \\ &=\frac{\frac{-3}{2}}{\frac{\sqrt{3}}{2}}\\&=-\frac{3}{\sqrt3}\\&=-\sqrt3\end{aligned}$

And since we know that R is (0, 2), R is the y-intercept of RM. Then, using the slope-intercept form:

$y=mx+b$

We can see that the equation of Line RM is:

$y=-\sqrt{3}x+2$

6 0

3 years ago

Read 2 more answers

Also they have positive 1.25

aleksley [76]

-1.25............................

6 0

3 years ago