Aerobics classes class $153.86 for 14 sessions. What is the fee for six sessions?

Ike plans to save the same amount in the 12th month as in each of the other months.

anyanavicka [17]

People who never develop symptoms or are not yet showing symptoms (pre-symptomatic) might not know that they are infected but can still spread rona to others. Masks also offer protection to the wearer.

6 0

3 years ago

Solve the following congruence equations for X a) 8x = 1(mod 13) b) 8x = 4(mod 13) c) 99x = 5(mod 13)

xxMikexx [17]

Answer:

a) 5+13k where k is integer

b) 20+13k where k is integer

c)12+13k where k is integer

Step-by-step explanation:

(a)

$8x \equiv 1 (mod 13) \text{ means } 8x-1=13k$ .

8x-1=13k

Subtract 13k on both sides:

8x-13k-1=0

Add 1 on both sides:

8x-13k=1

I'm going to use Euclidean Algorithm.

13=8(1)+5

8=5(1)+3

5=3(1)+2

3=2(1)+1

Now backwards through the equations:

3-2=1

3-(5-3)=1

3-5+3=1

(8-5)-5+(8-5)=1

2(8)-3(5)=1

2(8)-3(13-8)=1

5(8)-3(13)=1

So compare this to:

8x-13k=1

We see that x is 5 while k is 3.

Anyways 5 is a solution or 5+13k is a solution where k is an integer.

b)

$8x \equiv 4 (mod 13)$

8x-4=13k

Subtract 13k on both sides:

8x-13k-4=0

Add 4 on both sides:

8x-13k=4

We got this from above:

5(8)-3(13)=1

If we multiply both sides by 4 we get:

8(20)-13(12)=4

So x=20 and 20+13k is also a solution where k is an integer.

c)

[tex]99x \equiv 5 (mod 13)[/tex

99x-5=13k

Subtract 13k on both sides:

99x-13k-5=0

Add 5 on both sides:

99x-13k=5

Using Euclidean Algorithm:

99=13(7)+8

13=8(1)+5

Go back through the equations:

13-8=5

13-(99-13(7))=5

8(13)-99=5

99(-1)+8(13)=5

Compare this to 99x-13k=5 and see that x=-1 or -1+13=12 or 12+13k is a solution where k is an integer.

8 0

3 years ago

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What is the slope of the line that passes through the pair of points (1/3, -1) and ( -2, 9/2)

UkoKoshka [18]

Answer:

$\boxed{m = - \frac{33}{14} }$

.

Step-by-step explanation:

Use the form below

$\boxed{\boxed{m = \frac{y_2 - y_1}{x_2 - x_1} }}$

Where

$m$ is a slope
$(x_1,~y_1)$ and $(x_2,~y_2)$ are the point of the line

.

So, the slope is

$(\frac{1}{3} ,~ -1) \to x_1 = \frac{1}{3} ~and~y_1 = -1$

$(-2,~ \frac{9}{2} ) \to x_2 = -2~and~y_2 = \frac{9}{2}$

.

$m = \frac{\frac{9}{2} -(-1)}{-2-\frac{1}{3} }$

$m = \frac{\frac{11}{2} }{\frac{-7}{3} }$

$m = \frac{11}{2} \div (-\frac{7}{3} )$

$m = \frac{11}{2} \times (-\frac{3}{7} )$

$m = - \frac{33}{14}$

.

Happy to help:)

7 0

3 years ago

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. Imagine a game of 3 players where exactly one player wins in the end and all players have equal chances of being the winner. T

lbvjy [14]

Answer:

5/9

Step-by-step explanation:

Number of players = 3

number of times game is repeated = 4

P( any person wins a game ) = 1/3

P ( any person does not win a game ) = 1 - 1/3 = 2/3

P ( any person wins no game in 4 attempts ) = ( 2/3 )^4 = 16/81

Note : each player has equal chance of winning

Find the probability that there is at least one person who wins no games

lets represent the probability of each player not wining a game with alphabet A

A1 = player 1 wins no game

A2 = player 2 wins no game

A3 = players 3 wins no game

Applying the inclusion-exclusion formula

P( A1 ∪ A2 U A3 ) = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1 ∩ A3 ) + P( A1 ∩ A2 ∩ A3 )

where

P( A1 ∩ A2 ) = P( A1 wins all games )

P ( A1 wins all games in 4 attempts ) = ( 1/3 )^4 = 1/81

P( A1 ∩ A2 ∩ A3 ) = P ( no players wins any game in 4 attempts ) = 0

Hence

P( A1 ∪ A2 U A3 ) = 16/81 + 16/81 + 16/81 - 1/81 - 1/81 - 1/81 - 0 = 5/9

6 0

3 years ago

Plz help me with this

Jlenok [28]

Answer is 12 to your question

8 0

3 years ago

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