There are 6 girls and 7 boys in a class. A team of 10 players is to be selected from the class. How many different combinations of players are possible?]

Answers

Answer 1
Answer:

Answer:

In 286 different ways 10 players can be selected.

Step-by-step explanation:

There are 6 girls and 7 boys in a class. So in total there are 6+7 = 13 number of students in the class.

A team of 10 players is to be selected from the class.

As there is no other conditions are given, we can pick any 10 students from 13 students.

The way we can select 10 players from 13 students is,

=\dbinom{13}{10}

=(13!)/(10!(13-10)!)

=(13!)/(10!\ 3!)

=(13* 12* 11* 10!)/(10!\ 3!)

=(13* 12* 11)/(6)

=286

Answer 2
Answer: This is a simple combination problem. Just take the number of girls and boys, which is 13 and take 10 of it at a time. 13C10 is 286.


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The current (I) in an electrical conductor varies inversely as the resistance (R) of the conductor. The current is 6 amperes when the resistance is 722 ohms. What is the current when the resistance is 768 ohms? Round your answer to two decimal places if necessary.

Answers

Answer:

The current is 5.64 Ampere when resistance is  768 ohms .

Step-by-step explanation:

As given

The current (I) in an electrical conductor varies inversely as the resistance (R) of the conductor.

Thus

I \propto (1)/(R)

I=(k)/(R)

Where k is the constant of proportionality .

The current is 6 amperes when the resistance is 722 ohms.

I = 6 amperes

R = 722 ohms

Put all the values in the formula

6=(k)/(722)

k = 6 × 722

k = 4332

As given

when the resistance is 768 ohms .

R = 768 ohms

K = 4332

Put in the formula I=(k)/(R) .

I=(4332)/(768)

I = 5.64 Ampere

Therefore the current is 5.64 Ampere when resistance is  768 ohms .

Basically, this is Ohm's Law:
V=RI where V is the tension in volts;
You have that V=722×6=4332 Volts
Using this value for V you have that:
4332=768×I
So I=5.64 A

Am solving this problem correct

Answers

a) The degree of the first term is... 1
first \ term = 3t
The degree of the second term is... 2
second \ term = t^2
The degree of the third term is... 4
third \ term = t^4

b) The leading term of the polynomial is... 7t⁴
The leading coefficient of the polynomial is... 7

c) The degree of the polynomial is... 4
a) 1, 2, 4
b) 7t^4,7
c) 4

a long distance telephone company charges a rate of 8 cents per minute or 90 cent mininum charge per call, whichever is greater. Find the cost in cents per minute for a 41 minute call.

Answers

The cost of a 41 min call would be 328 cents. So, that would be 8 cents per minute, because the charge of the call is greater that 90 cents. Hope that helps!!

What are the next four multiples of the fraction 2/6

Answers

4/12 6/18 and 8/24 i multiplied them by 2 3 and 4

Answer:

yo mommy

Step-by-step explanation:

Which table is a justification for the given solution to the system of equations?(2, −1)
(2, −1)


y={4x−92x−5

Answers

Given that y = 4x – 9 and y = 2x - 5 (2 , -1)
substituting equation 1 to the equation 2  
Y = 4x – 9
4x – 9 = 2x - 5
2x = 4
x = 2    
Substitute 2  
Y = 2(2) - 5
Y = -1
So the ordered pair a the solution the system of equtions given

Which strategie would eliminate a variable in the system of equations? −x+6y=8 7x−y=−2 ​

Answers

Answer:

  substitution (or addition)

Step-by-step explanation:

A simple strategy for this system is to use substitution. The first equation is easily solved for x, so you could substitute that into the second equation:

  x = 6y -8

  7(6y -8) -y = -2 . . . . . x variable eliminated

__

The second equation is easily solved for y, so you could substitute that into the first equation.

  y = 7x +2

  -x +6(7x +2) = 8 . . . . . y-variable eliminated

__

The "addition" method is always a good way to eliminate a variable.

When the coefficient of a variable in one equation is a divisor of the coefficient of that variable in the other equation, a simple multiplication and addition will do.

To make the coefficient of x in the first equation the opposite of the coefficient of x in the second, multiply the first equation by 7. Adding that result to the second equation will eliminate x:

   7(-x +6y) +(7x -y) = 7(8) +(-2)

  42y -y = 56 -2 . . . . . . x-variable eliminated

Likewise, the second equation can be multiplied by 6 and added to the first to eliminate the y-variable:

  (-x +6y) +6(7x -y) = (8) +6(-2)

  -x +42x = -4 . . . . . . . . y-variable eliminated

__

It is often the case that using either substitution or "addition" requires about the same amount of work.

Here, the solutions are (x, y) = (-4/41, 54/41).

Final answer:

To eliminate a variable in the given system of equations, you can use the elimination method. By multiplying the equations by suitable numbers and adding them, you can cancel out one of the variables, simplifying the process to solve for the other variable.

Explanation:

You can eliminate a variable in the given system of equations: −x+6y=8 and 7x-y=−2 by using either the substitution method or the elimination method. For this scenario, the elimination method will work best.

Strategy:

  1. To eliminate x, you should first multiply the first equation by 7 and the second by 1, resulting in the equations: -7x+42y=56 and 7x-y=-2
  2. Adding these two equations together, the x terms (-7x and 7x) cancel out, giving us: 41y=54.
  3. Finally, you divide both sides by 41 to solve for y. This process effectively eliminates the variable x from the equation, providing a solution for y.

This variable eliminationstrategy lets you solve one equation for one variable, simplifying the process of finding solutions for a system of equations.

Learn more about Variable Elimination here:

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