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PtichkaEL [24]
3 years ago
15

The composer Beethoven wrote 9 symphonies, 5 piano concertos (music for piano and orchestra), and 32 piano sonatas (music for so

lo piano). a. How many ways are there to play first a Beethoven symphony and then a Beethoven piano concerto?b. The manager of a radio station decides that on each successive evening (7 days per week), a Beethoven symphony will be played followed by a Beethoven piano concerto followed by a Beethoven piano sonata. For how many years could this policy be continued before exactly the same program would have to be repeated?
Mathematics
1 answer:
Gekata [30.6K]3 years ago
4 0

Answer:

a) 45 ways

b) 3.95 years

Almost 4 years!

Step-by-step explanation:

The problem of how many ways different selections can be made with order important is a permutation and combination problem

a) The number of ways there are to play first a Beethoven symphony and then a Beethoven piano concerto.

There are 9 Beethoven symphonies to select from.

There are 5 Beethoven piano concertos to select from.

The total number of ways the selection can be made = ⁹C₁ × ⁵C₁ = 45 ways.

b) How many ways to play a Beethoven symphony, followed by a piano concerto and then a piano sonata.

There are 9 Beethoven symphonies to select from.

There are 5 Beethoven piano concertos to select from.

There are 32 Beethoven piano sonatas to select from.

Total number of ways = ⁹C₁ × ⁵C₁ × ³²C₁ = 1440 ways.

With each day, taking each way, this means, it'll take 1440 days to exhaust the different number of ways to do this

There are 7 days in a week,

So, 1440 days has (1440/7) weeks; 205.71 weeks

52 weeks = 1 year

205.71 weeks = (205.71/52) = 3.95 years

Almost 4 years!

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James cut out four parallelograms, the dimensions of which are shown below. Parallelogram 1 length: 12 in. width: 15 in. diagona
MrRa [10]

Answer: The correct options are;

*The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 1 do not form right angles

**The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 4 do not form right angles.

Step-by-step explanation: What James is trying to do is quite simple which is, he wants to place four quadrilaterals inside a circle and he wants the vertices to touch one another at the center of the circle without having to overlap.

This is possible and quite simple, provided all the quadrilaterals have right angles (90 degrees). This is because the center of the circle measures 360 degrees and we can only have four vertices placed there without overlapping only if they all measure 90 degrees each (that is, 90 times 4 equals 360).

We can now show whether or not all four parallelograms have right angles by applying the Pythagoras' theorem to each of them. Note that James has cut the shapes in such a way that the hypotenuse (diagonal) and the other two legs have already been given in the question. As a reminder, the  Pythagoras' theorem is given as,

AC² = AB² + BC² Where AC is the hypotenuse (diagonal) and AB and BC are the other two legs. The experiment would now be as follows;

Quadrilateral 1;

20² = 12² + 15²

400 = 144 + 225

400 ≠ 369

Therefore the vertices of parallelogram 1 do not form a right angle

Quadrilateral 2;

34² = 16² + 30²

1156 = 256 + 900

1156 = 1156

Therefore the vertices of parallelogram 2 forms a right angle

Quadrilateral 3;

29² = 20² + 21²

841 = 400 + 441

841 = 841

Therefore the vertices of parallelogram 3 forms a right angle

Quadrilateral 4;

26² = 18² + 20²

676 = 324 + 400

676 ≠ 724

Therefore the vertices of parallelogram 4 do not form a right angle

The results above shows that only two of the parallelograms cut out have right angles (like a proper square or rectangle for instance), while the other two do not have right angles.

Therefore, the correct option are as follows;

The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 1 do not form right angles.

The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 4 do not form right angles.

8 0
3 years ago
Read 2 more answers
Pls help plssadasdsadsadasd
EleoNora [17]

Answer:

option B

hope its helpful

make me brainliest plz

sorry if it isn't correct

6 0
3 years ago
NEED HELP ON THIS QUESTION ASAP
Katarina [22]

Answer:

Step-by-step explanation:

A+B+C+D=360

A=180 as it half of the circle

B+C+D=180

60+C+90=180

150+C=180

C=180-150=30

The fraction of the circle to region C

30/180=1/6

8 0
4 years ago
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For question 5 plz :)
RideAnS [48]

Answer: b

Step-by-step explanation: i took the quiz

8 0
4 years ago
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What is the area of this figure? Use 3.14 for pi.
andrey2020 [161]

Answer:

24,53cm²

Step-by-step explanation:

rectangle area=length×width=7×3

=21cm²

circle area=1/2Π r²

=1/2×3.14×1.5×1.5

=3.5325cm²

add the two areas

21+3.53=24,53cm²

6 0
3 years ago
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