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Lady bird [3.3K]
3 years ago
9

Select True Or False For Each Statement. Reward: Brainliest

Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
4 0
Answers are in an attached photo

Gre4nikov [31]3 years ago
3 0
False, true, true, false
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Find the are of the SHADED region.
Mazyrski [523]
First find the area of the parallelogram


Area of parallelogram = b x h
= 12 x 7
= 84

Then find area of kite

Area of kite = (xy)/2
= (7 x 12)/2
= 84/2
= 42

Now we are gonna subtract the area of the kite from the area of the parallelogram.

84 - 42 = 42

Therefore the area of the shaded region is 42.
7 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
lng rordoin a party contextat the Vine House HotelYour House Hotelparty $20 $750 Chargesfor Sarah per 30 ow thperson pluspeoplet
S_A_V [24]

So in this problem we can see that there will be 60 people at the party. Following the conditions for this event that state the following: $750 for up to 30 people and every extra person pays $20. To show that the cost will be $1350 in total we simply do this:



For 30 people we have $750 dollars. Now every other person after pays only $20. That means that another the other 30 people will pay in total only $600.



If we add them we get: $750 + $600 = 1350. This shows that the cost will be $1350.

<span>
I hope it helps, Regards.</span>

3 0
3 years ago
HELLllllllPPppppp !!!
Tasya [4]

9514 1404 393

Answer:

 BC ≈ 17.0 (neither Crow nor Toad is correct)

Step-by-step explanation:

The left-side ratio of (2+4)/4 = 3/2 suggests BC is 3/2 times the length DE. If that were the case, BC = (3/2)(11) = 16.5, as Crow says.

The right-side ratio of (5+9)/9 = 14/9 suggests that BC 9 is 14/9 times the length DE. If that were the case, BC = (14/9)(11) = 154/9 = 17 1/9 ≈ 17.1, as Toad says.

The different ratios of the two sides (3/2 vs 14/9) tell you that the triangles are NOT similar, so the length of BC cannot be found by referring to the ratios of the given sides.

Rather, the Law of Cosines must be invoked, first to find angle A (109.471°), then to use that angle to compute the length of BC given the side lengths AB and AC. That computation gives BC ≈ 16.971. (See the second attachment.)

5 0
3 years ago
Helppppp<br>needdddd it​
Olegator [25]

Answer:

angle don+angle gil=90°

don+ 74°=90°

don=90°-74°

don=16°

angle e + angle a=180°

128°+angle a=180°

angle a=180°-128°

angle a=52°

8 0
3 years ago
Read 2 more answers
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