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3 years ago

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A classroom of children has 18 boys and 19 girls in which five students are chosen to do presentations. in how many ways can the

five students be chosen so that more boys than girls are selected?

1 answer:

il63 [147K]3 years ago

7 0

They can be chosen in 12,444 ways.

There are three different scenarios that have more boys being chosen than girls:
5 boys and 0 girls
4 boys and 1 girl
3 boys and 2 girls

We will set up combinations for all 3 of these scenarios:
$(_{18}C_5\times{_{19}C_0)+(_{18}C_4\times_{19}C_1)+(_{18}C_3\times_{19}C_2)$
$\frac{18!}{5!13!}\times \frac{19!}{0!19!}+\frac{18!}{4!14!}\times \frac{19!}{1!18!}+\frac{18!}{3!15!}\times \frac{19!}{2!17!}
\\
\\=8568+3060+816=12444$

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Answer:

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3 years ago

PLEASE HELP!!!!

Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

if "a" is positive, then opens UP
if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

the Axis Of Symmetry (x-value) is: x = $\frac{-b}{2a}$
y-value is found by plugging in the AOS for "x" in the equation

****************************************************************************************

1) y = x² + 11x + 24

a = +1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-11}{2(1)}$ = $-\frac{11}{2}$

y = $(-\frac{11}{2})^{2}$ + 11 $(-\frac{11}{2} )$ + 24 = $-\frac{25}{4}$

vertex $(-\frac{11}{2}$ , $-\frac{25}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

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2) y = -x² - 6x - 8

a = -1 so the parabola opens DOWN

x = $\frac{-b}{2a}$ = $\frac{-(-6)}{2(-1)}$ = $\frac{6}{-2}$ = -3

y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1

vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

******************************************************************************************

3) y = x² - 2x + 3

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-(-2)}{2(1)}$ = $\frac{2}{2}$ = 1

y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2

vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). <em>There are no x-intercepts</em>

******************************************************************************************

4) y = x² + 4x + 4

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-4}{2(1)}$ = -2

y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0

vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). <em>The x-intercept is the vertex.</em>

Compared to the other four graphs, this is most likely the equation for the rain gauge!

******************************************************************************************

5) y = 3x² + 21x + 30

a = +3 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-21}{2(3)}$ = $-\frac{7}{2}$

y = 3 $(-\frac{7}{2})^{2}$ + 21 $(-\frac{7}{2} )$ + 30 = $-\frac{27}{4}$

vertex $(-\frac{7}{2}$ , $-\frac{27}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

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4 0

3 years ago

Read 2 more answers

PLEEAASEEE HELP! what is the area of trapezoid ABCD?

daser333 [38]

What is the area of trapezoid ABCD ?

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3 0

3 years ago

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Dont answer this PLEAS

pickupchik [31]

Answer:

why? also the answer is 1.6y+y-4/15y+7/6y=21/3

We move all terms to the left:

1.6y+y-4/15y+7/6y-(21/3)=0

Domain of the equation: 15y!=0

y!=0/15

y!=0

y∈R

Domain of the equation: 6y!=0

y!=0/6

y!=0

y∈R

We add all the numbers together, and all the variables

1.6y+y-4/15y+7/6y-7=0

We add all the numbers together, and all the variables

2.6y-4/15y+7/6y-7=0

We calculate fractions

2.6y+(-24y)/90y^2+105y/90y^2-7=0

We multiply all the terms by the denominator

(2.6y)*90y^2+(-24y)+105y-7*90y^2=0

We add all the numbers together, and all the variables

(+2.6y)*90y^2+(-24y)+105y-7*90y^2=0

We add all the numbers together, and all the variables

105y+(+2.6y)*90y^2+(-24y)-7*90y^2=0

Wy multiply elements

-630y^2+105y+(+2.6y)*90y^2+(-24y)=0

We get rid of parentheses

-630y^2+105y+(+2.6y)*90y^2-24y=0

We add all the numbers together, and all the variables

-630y^2+81y+(+2.6y)*90y^2=0

Step-by-step explanation:

5 0

3 years ago

Hello, help me please)

AlladinOne [14]

Answer:

Given equation:

$10^{5x-2}=2^{8x-3}$

Take natural logs of both sides:

$\implies \ln 10^{5x-2}= \ln 2^{8x-3}$

$\textsf{Apply log Power law}: \quad \ln_ax^n=n\ln_ax$

$\implies (5x-2)\ln 10=(8x-3) \ln 2$

Expand brackets:

$\implies 5x\ln 10 - 2\ln 10=8x \ln 2 -3 \ln 2$

Collect like terms:

$\implies 5x\ln 10 - 8x \ln 2 =2\ln 10-3 \ln 2$

Factor left sides:

$\implies x(5\ln 10 - 8 \ln 2) =2\ln 10-3 \ln 2$

$\textsf{Apply log Power law}: \quad \ln_ax^n=n\ln_ax$

$\implies x(\ln 10^5 - \ln 2^8) =\ln 10^2- \ln 2^3$

$\textsf{Apply log Quotient law}: \quad \ln_a\frac{x}{y}=\ln_ax - \ln_ay$

$\implies x\left(\ln\left(\dfrac{10^5}{2^8}\right)\right) =\ln\left(\dfrac{10^2}{2^3}\right)$

Simplify:

$\implies x\left(\ln\left(\dfrac{3125}{8}\right)\right) =\ln\left(\dfrac{25}{2}\right)$

$\implies x=\dfrac{\ln\left(\dfrac{25}{2}\right)}{\ln\left(\dfrac{3125}{8}\right)}$

$\implies x=0.4232297737...$

6 0

2 years ago