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Vsevolod [243]
3 years ago
12

At Mountain High School, the students were surveyed about their participation in band (B) and track (T). The results of the surv

ey are shown in the Venn diagram. Given that a randomly chosen student participates in band, what is the probability that the student also participates in track?
Mathematics
2 answers:
mixer [17]3 years ago
7 0
The Venn diagram is not included, so I am going to explain the procedure and you will be able to calculate the result just plugging the numbers.

From Vend diagram you will be able to conclude two numbers:

1) the total number os students surveyed that participate in band (B)

2) the total number of students surveyed that participate in track (T)

3) the total number of students surveyed that participate in both band (B) and track (T) = B∩T

4) ) the total number of students surveyed, which is: total number of students in band B + total number of students in track - less the number of students that participate in both. = B + T - (B∩T)

So, the probability that one student that participate in band (B) also participate in track (T) =

Number of students that participate in both / number of students that participate in band = (B∩T) / B

Now, you just have to find B∩T and B from the diagram and plugg in into the formula (B∩T)/B. That's it.

borishaifa [10]3 years ago
4 0
E2020 Answer is : B (9/33)
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Can someone please help with this or explain how to do it. I'm not very sure where to begin ToT
larisa86 [58]

Answer:

d. 12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)

Step-by-step explanation:

First, we find the greatest common factor of 12x^23y^40z^9 and 10x^78z^5.

We work on the numbers, then on x, then on y, then on z.

We start with the GCF of 12 and 10.

10 = 2 * 5

12 = 2 * 3 * 3

The only factor 10 and 12 have in common is 2, so the GCF of 12 and 10 is 2.

Now we work on x. We have x^23 and x^78.

x^23 has 23 factors of x.

x^78 has 78 factors of x.

Since x^23 and x^78 have both at least 23 factors of x, x^23 is the GCF of x.

Now we work on y.

There is y^40 in one term, but there is no y in the other term. There is no GCF for the y.

Now we work on z. The first term has z^9. the second term has z^5. They both have at least 5 factors of z, so the GCF of z is z^5.

Now we put all the parts of the GCF together, and the GCF of the two terms is

2x^23z^5

We use option d. to factor.

12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(___a____ + ____b___)

We already have the GCF outside the parentheses. Now we need to figure out what goes inside the parentheses.

The terms that go in positions a and b are the terms that when multiplied by the GCF give you the original expression.

We now work on position a.

What do you multiply by 2x^23z^5 to get 12x^23y^40z^9?

Answer: 6y^40z^4

Now we have

12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + ____b___)

Now we do the same for position b.

What do you multiply by 2x^23z^5 to get 10x^78z^5?

Answer: 5x^55

Now we have

12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)

Answer: d. 12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)

5 0
3 years ago
Consider the graph shown.
otez555 [7]

Answer:

y = \frac{2}{3} x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (0, 2) and (x₂, y₂ ) = (3, 4)

m = \frac{4-2}{3-0} = \frac{2}{3}

The line crosses the y- axis at (0, 2) ⇒ c = 2

y = \frac{2}{3} x + 2 ← equation of line

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timofeeve [1]

Answer:

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Step-by-step explanation:

For the perimeter of a shape you would add all the sides together. So in this case it would be:

7mm+7mm+7mm+7mm+7mm+7mm=42mm

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Answer:
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