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2 years ago

A class consists of 63 women and 86 men. If a student is randomly selected, what is the probability that the student is a woman?

2 answers:

6 0

First, we need to get the total number students by adding the number of men and women. 63 + 86 = 149. This is the total number of samples. To get the probability that a randomly selected is a woman, we can write the ratio as 63:149 or the probability is 63/149.

VikaD [51]2 years ago

5 0

Answer:

63/149

Step-by-step explanation:

just took the test on edge.

Hope this helps❣️

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2 years ago

Y + 4x = 8 and 5x + 2y = 13

Ivanshal [37]

Y + 4x = 8
y = 8 - 4x
substitute 8 - 4x for y in the other equation.
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3 years ago

Help please it would help my grade alot

Alexxandr [17]

Answer:

It cooled 58.4 degrees

Step-by-step explanation:

T(t) = 68 + 144e ^(-.052t)

We want to find the value when t= 10

T(10) = 68 + 144e ^(-.052*10)

= 68 +85.611

T(10) =153.611

We also need to know

T(0)= 68 + 144e ^(-.052*0)

=68+144

=212

We want to know how much it cools

That would be the temperature at time 0 minus the temperature at time 10

T(0) - T (10)

212-153.611

58.389

To the nearest tenth

58.4 degrees

3 0

2 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

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