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4 years ago

How do you solve this problem? Thank you!

Mathematics

1 answer:

Andrej [43]4 years ago

7 0

If im not mistaken, you would use the Pythagorean theorem.

So, its c^2= A^2 + b^2

So, 10^2 = 1x^2 + 3x^2

100 = 4x^2 (You can add the x's since they are both x and both to the power of 2)

I would then use a square root to get rid of that power of two

The square root of 100 is 10, and the square root cancels out on the other side.

10 = 4x

10/4 = 2.5

x = 2.5

I hope that helps!

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2 years ago

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antoniya [11.8K]

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3 years ago

In a school of 450 people 110 are in the choir 240 are in a band and 60 are in both

zhannawk [14.2K]

The question doesn't require any specific probabilities, but I'm adding my own calculations to make it easier for you to solve your own problem

Answer:

Questions added and answered below

Step-by-step explanation:

Venn Diagram

When we have different sets, some of them belong only to one set, some belong to more than one, some don't belong to any of them. This situation can be graphically represented by the Venn Diagrams.

Let's analyze the data presented in the problem and fill up the numbers into our Venn Diagram. First, we must use the most relevant data: there are 60 people in both the choir and the band. This number must be in the common space between both sets in the diagram (center zone, purple).

We know there are 110 people in the choir, 60 of which were already placed in the intersection zone, so we must place 110-60=50 people into the blue zone, belonging to C but not to B.

We are also told that 240 people are in a band, 60 of which were already placed in the intersection zone, so we must place 240-60=180 people into the red zone, belonging to B but not to C.

Finally, we add the elements in all three zones to get all the people who are in the choir or in the band, and we get 50+60+180=290. Since we have 450 people in the school, there are 450-290=160 people who are not in the choir nor in the band.

The question doesn't ask for a particular probability, so I'm filling up that gap with some interesting probability calculations like

a) What is the probability of selecting at random one person who is in the band but not in the choir?

The answer is calculated as

$\displaystyle P(A)=\frac{180}{450}=0.4$

b) What is the probability of randomly selecting one person who belongs only to one group?

We look for people who are in only one of the sets, they are 50+180=230 people, so the probability is

$\displaystyle P(B)=\frac{230}{450}=0.51$

b) What is the probability of selecting at random one person who doesn't belong to the choir?

We must add the number of people outside of the set C, that is 180+160=340

$\displaystyle P(C)=\frac{340}{450}=0.76$

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3 years ago