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aivan3 [116]
3 years ago
8

Does the trapizoid have at lease 1 pair of parallel sides

Mathematics
2 answers:
Artyom0805 [142]3 years ago
4 0
All trapezoid has 1 parallel line on the opposite side so I suggest that it does have a pair of parallel sides. I Hope I HELPED YOU :D
mezya [45]3 years ago
3 0
All tapezoids have 1 parallel side
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Three consecutive integers have a sum of 72 . Find the integers.
ANTONII [103]

Answer:

23, 24 and 25

Step-by-step explanation:

These are consecutive numbers that add up to 72.

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3 years ago
Lynn wants to serve hamburgers at her birthday party if hamburger buns are sold in packages of 8 and hamburger patties are sold
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3 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
How do you solve this? I really don't understand...
maksim [4K]

Answer:

MEAN 5.8 add all the points up then divide by the number of dots

MEDIAN 5 the middle of the data

MODE 5 most accruing one

RANGE 7 maximum minus minimum

8 0
3 years ago
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