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Aleksandr-060686 [28]
3 years ago
7

Hockey teams receive 2 points when they win and 1 point when they tie. One​ season, a team won a championship with 66 points. Th

ey won 12 more games than they tied. How many wins and how many ties did the team​ have?
Mathematics
1 answer:
german3 years ago
3 0

Answer:

26 wins , 14 ties

Step-by-step explanation:

Given: Hockey teams receive 2 points when they win and 1 point when they tie. One​ season, a team won a championship with 66 points. They won 12 more games than they tied.

To Find: How many wins and how many ties did the team​ have.

Solution:

let the number of wins hockey team have =\text{x}

let the number of ties hockey team have   =\text{y}

Now,

Hockey teams receive 2 points when they win and 1 point when they tie.

2\text{x}+\text{y}=66

They won 12 more games than they tied

\text{x}-\text{y}=12

Solving both equations we get

\text{x}=26

\text{y}=14

Hence, the hockey team have 26 wins and 14 ties.

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8 0
4 years ago
A card is being drawn from a standard deck of playing cards. Find each probability.
EastWind [94]
These are 6 questions and 6 answers.

To find each probability we will use the definition of probability:

Probability = number of positive outcomes / number of total possible outcomes

1) <span>P(Jack or ten)
</span>

<span>Answer: 2/13 ≈ 0.12
</span>

Justification:

i) Positive outcomes: A standard deck of cards has 4 jacks and 4 tens, then those are 4 + 4 = 8 different positive outcomes.

ii) Possible outcomes: a standard deck of cards has 52 different cards, so, that is a total of 52 different possible outcomes

 
iii) Probability, P


P = number of positive outcomes / number of total possible outcomes

P = 8 / 52 = 2/13 ≈ 0.15

<span> 2.P(red or black)
</span>

Answer: 1

Justification:

i) Positive outcomes

Half of the cards are red and half of the cards are black, so they both add for the total of the cards = 52


ii) Possible outcomes: 52 cards


iii) Probaility, P 

P = number of positive outcomes / number of total possible outcomes

P = 52 / 52 = 1

<span> 3.P(queen or club)
</span>

Answer: 4/13 ≈ 0.31

Justification:


i) Positive outcomes

There are 4 Queens.

There are 1/4 of 52 clubs = 1/4 × 52 = 13 clubs.

But you cannot add all of them, because one club is the Quenn of Clubs.

Then, the total number of different Queens and clubs is 4 + 13 - 1 = 16



ii) Possible outcomes: 52 different cards


iii) Probaility, P

P = number of positive outcomes / number of total possible outcomes

P = 16 /52 = 4 / 13 ≈ 0.31


<span> 4.P(red or ace)
</span>

Answer: 7 / 13 ≈ 0.54


Justification:


i) Positive outcomes

Half of the cards are red: 26

There are 4 aces. 

Since 2 aces are red, the number of different red and aces cards is: 26 + 4 - 2 = 28

ii) Possible outcomes: 52 different outcomes


iii) Probaility, P

P = number of positive outcomes / number of total possible outcomes

P = 28 / 52 = 7 / 13 ≈ 0.54

<span> 5.P(diamond or black)
</span><span>
</span>
Answer: 1/2 = 0.5

Justification:

i) Positive outcomes

There are 52 / 4 = 13 diamonds

There are 26 black cards.


All the diamonds are black cards.

Then, the number of different diamond or black cards is 13 + 26 - 13 = 26

ii) Possible outcomes: 52 different cards.

iii) Probaility, P

P = number of positive outcomes / number of total possible outcomes

P = 26 / 52 = 1/2 = 0.5

6.P(face card or spade)

Answer: 11/26 ≈ 0.42


Justification:

i) Positive outcomes

Face cards are jacks, queens and kings. That is 3 × 4 = 12 different cards.

The spades are 13 cards.

Since, 3 of the faces are spade cards, the number of different cards of those types are 12 + 13 - 3 = 22


ii) Possible outcomes: 52 different cards


iii) Probaility, P

P = number of positive outcomes / number of total possible outcomes

P = 22 / 52 = 11 / 26 ≈ 0.42
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Jet001 [13]

Answer:

The roots are x = (\frac{3}{2}, -2), which is given by option C.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}

x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}

\Delta = b^{2} - 4ac

In this question, we have that:

2x^2 + x - 6 = 0

Which is a quadratic equation with a = 2, b = 1, c = -6. So

\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49

x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}

x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2

So the roots are x = (\frac{3}{2}, -2), which is given by option C.

3 0
3 years ago
Please help me this is due in a few minutes
Vaselesa [24]

Answer:

Volume of one Bugle => 462.8 mm³

Volume of 18 Bugles => 8,330.4 mm³

Volume formula for a cone => ⅓πr²h

Step-by-step explanation:

Volume of one bugle = volume of cone (V) = ⅓πr²h

Where,

r = 5.1 mm

h = 17 mm

Volume of one Bugle = ⅓*π*5.1²*17 = 462.8 mm³

Volume of 18 Bugles = 462.8 × 18 = 8,330.4 mm³

7 0
3 years ago
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