Ask question

Ask question

All categories

4 years ago

5

In how many ways can 4 different contracts be distributed among 12 different firms, if no two contracts are awarded to the same

firm?

1 answer:

vagabundo [1.1K]4 years ago

5 0

The first contract can go to any of 12 firms. The second can go to any of the remaining 11 firms, and so on. Finally, the 4th contract can go to any of the 9 firms that don't yet have a contract.

The number of ways the contracts can be awarded is

... 12×11×10×9 = 11,880

You might be interested in

Given n//p m<2=50 degrees what is m<7?

Bogdan [553]

Answer: m∠7=130°

Step-by-step explanation:

If two parallel lines are cut by another line (transversal), each one will have four angles surrounding the intersection.

By definition, the adjacent angles of each parallel line, are supplementary, this means that they add up 180°.

Then, if m∠2=50°, then m∠3 is:

m∠3 $=180-50=130$ °

By definition, the angles that are in the same relative position are know as corresponding angles and they are congruent.

Therefore:

m∠3=m∠7

m∠7=130°

3 0

3 years ago

Read 2 more answers

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

Can someone please help me out with my grades are bad

dolphi86 [110]

Answer:

Step-by-step explanation:

In a dilatation for find the new coordinates we have to multiply the initial value for the the scale factor

A’(-3, 12)

B’(9,6)

C’(-9,-3)

so the general rule is:

(x,y) -> (3x, 3y)

7 0

3 years ago

In a wrestling contest, a wrestler needs a winning percentage of at least 70% to qualify for round 2. sam lost 3 out of 12 match

algol [13]

He won about 25% because 6/12 would be 1/2 and 3 is 1/2 of 6 or a 1/4 so 25%

8 0

4 years ago

I would gladly appreciate it

IrinaVladis [17]

Answer: D

Step-by-step explanation: Complementary angles add up to 90 degrees supplementary add up to 180 degrees and vertical angles are like CAF and DAE like how they're diagonal

7 0

3 years ago

Read 2 more answers