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3 years ago

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In how many ways can a committee of 3 men and 4 boys be chosen from 7 men and 6 boys so as not to include the youngest boy if th

e eldest man is serving?

1 answer:

Ludmilka [50]3 years ago

6 0

Answer:

There are 75 ways to form the committee.

Step-by-step explanation:

The order in which the people are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Considering the eldest has to be there, 2 men from a set of 6 and 4 boys from a set of 5(excluding the youngest), so:

$T = C_{6,2}C_{5,4} = \frac{6!}{2!4!} \times \frac{5!}{1!4!} = 3*5*5 = 75$

There are 75 ways to form the committee.

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the Johnsons want to cover their backyard with new grass their backyard is rectangular with a length of 3x - 2 feet and a width

Andre45 [30]

<h2>Hello!</h2>

The answer is:

The area that the want to cover with new grass will be:

$Area_{grass}=8x^{2} -12x-22$

<h2>Why?</h2>

If we want to calculate what is the ara of the region of the backyard that they want to cover with new grass, we need to calculate the area of the backyard, calculate the area of the swimming pool and its patio, and the last step is to subtract the area of the swimming pool and its patio to the total area of the backyard.

So, to calculate the areas, we need to use the following formula:

$Area=Length*Width$

Also, we need to remember how the distributive property works, since we are going to need it.

We have that:

$(a+b)(c+d)=ac+ad+bc+bd$

We are given the following information:

For the backyard we have:

$Length=3x-2\\Width=3x+1$

For the swimming pool and its patio:

Since there it's not mentioned, let be the first expression the length, and the second one, the width.

$Length=x+4\\Width=x+5$

Then, calculating we have:

Backyard:

$Area_{backyard}=(3x-2)*(3x+1)=9x^{2}+3x-6x-2=9x^{2}-3x-2$

Swimming pool and patio:

$Area_{swimmingpool}=(x+4)*(x+5)=x^{2}+5x+4x+20=x^{2}+9x+20$

Now, calculating the area that they want to cover with new grass, we have:

$Area_{grass}=Area_{backyard}-Area_{swimmingPool}\\\\Area_{grass}=(9x^{2}-3x-2)-(x^{2}+9x+20)\\\\Area_{grass}=9x^{2} -x^{2} -3x-9x-2-20\\\\Area_{grass}=8x^{2} -12x-22$

Hence, we have that the area that the want to cover with new grass will be:

$Area_{grass}=8x^{2} -12x-22$

Have a nice day!

3 0

3 years ago

Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

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3 years ago

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galina1969 [7]

Answer:

-4

Step-by-step explanation:

given

f(x) = 4x - 12

if x = 2,

f(2) = 4(2) - 12

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4 years ago

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Part A. Let $f(x) = x^4-3x^2 + 2$ and $g(x) = 2x^4 - 6x^2 + 2x -1$. Let $a$ be a constant. What is the largest possible degree o

Varvara68 [4.7K]

Answer:

A. 4

B. 1

Step-by-step explanation:

The degree of a one-variable polynomial is the largest exponent of the variable.

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<h3>A.</h3>

For f(x) = x^4 -3x^2 +2 and g(x) = 2x^4 -6x^2 +2x -1, the sum f(x) +a·g(x) will be ...

(x^4 -3x^2 +2) +a(2x^4 -6x^2 +2x -1)

= (1 +2a)x^4 +(-3-6a)x^2 +2ax -a

The term with the largest exponent is (1 +2a)x^4, which has degree 4. This term will be non-zero for a ≠ -1/2.

The largest possible degree of f+ag is 4.

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<h3>B.</h3>

The polynomial sum is ...

f+bg = (1 +2b)x^4 +(-3-6b)x^2 +2bx -b

When b = -1/2, the first two terms disappear and the sum becomes ...

f+bg = -x +1/2 . . . . . . a polynomial of degree 1

The smallest possible degree of f+bg is 1.

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2 years ago

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Paraphin [41]

Answer:

15.072

Step-by-step explanation:

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3 years ago

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