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DENIUS [597]
4 years ago
7

Whatever the answer to. -7y=28

Mathematics
2 answers:
Arlecino [84]4 years ago
3 0

Answer:

y=-24

Step-by-step explanation:

ziro4ka [17]4 years ago
3 0

Answer:

y = -4

Step-by-step explanation:

to find that you have to isolate the variable, y

and since -7 is being multiplied to the y you have to get rid of it by doing the opposite, dividing

and whatever you do to one side of the equal sign you MUST do it to the other

so

-7y = 28

/-7     /-7

y = -4

and there's your answer

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In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
The midsegment of ABC is LM. What is the length of AB if LB is 14 inches long​
morpeh [17]

Answer:

AB = 28 inches

Step-by-step explanation:

The length of LB = 14 inches

But; LB = AL

Therefore; AL = 14 inches

But; AB = AL + LB

Thus; AB = 14 inches + 14 inches

               = 28 inches

8 0
4 years ago
Read 2 more answers
Since I can't delete my previous question, can someone help me with these problems? I couldn't edit my last question so I am put
Marta_Voda [28]
XDXD you should be able to edit it
8 0
3 years ago
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Point P lies on circle C, x^2+y^2=36. There are two unique tangents to circle D, x^2+y^2=18 that pass through point p. What is t
Anastasy [175]

Answer:

90°

Step-by-step explanation:

Circle C has radius √36 = 6

Circle D has radius √18 = 3√2

As the D radius intersecting either point of tangency, does so at a right angle.

The half angle θ between the tangents is

sinθ = 3√2 / 6 = ½√2

θ = 45°

so the angle between the tangents is 2(45) = 90°

6 0
3 years ago
8x^3=(-2x)(A) what does a equal HELPPP PLEASERE
nikdorinn [45]

Answer:

a=0

Step-by-step explanation:

7 0
3 years ago
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