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Anni [7]
3 years ago
15

1/8 the sum of 23 and 17

Mathematics
1 answer:
dexar [7]3 years ago
6 0

Answer:

5 is the right answer.......

1/8 (23+17) =5

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What is (w-4/9)(-2/3)=-4/5
pshichka [43]
<span>(-2/3)*(w-(4/9)) = -4/3 // + -4/3

(-2/3)*(w-(4/9))-(-4/3) = 0

(-2/3)*(w-4/9)+4/3 = 0

4/3-2/3*(w-4/9) = 0

(-2/3*3*(w-4/9))/3+4/3 = 0

4-2/3*3*(w-4/9) = 0

44/9-2*w = 0

(44/9-2*w)/3 = 0

(44/9-2*w)/3 = 0 // * 3

44/9-2*w = 0

44/9-2*w = 0 // - 44/9

-2*w = -44/9 // : -2

w = -44/9/(-2)

w = 22/9

w = 22/9</span>
8 0
3 years ago
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
50 cases in 44 minutes equals how many cases in 60 minutes
Blizzard [7]
To solve this, set up a proportion, crossmultiply, and solve for x. 

Let x = the unknown amount of cases in 60 minutes.

\frac{50}{44} = \frac{x}{60}
44x = 50(60)
44x = 3000

Divide both sides by 44 to isolate variable x
x = 3000 / 44
x ≈ 68.1818 (the 18s repeat forever)

Rounding to the closest case and assuming the rate stays constant, you would get approximately 68 cases in 60 minutes. If you need not round to the nearest case, you would get approximately 68.18 cases in 60 minutes. 
4 0
3 years ago
What is 4/5 plus 2/10 plus 6/20
Elza [17]
<u>1 3/10 Because 4/5+2/10= 1 plus 6/20 [3/10] is 1 3/10</u>
6 0
4 years ago
Read 2 more answers
1. What is the degree form for 4π/9
nikklg [1K]
the answer should be about 80 degrees
8 0
4 years ago
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