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3 years ago

16. What statement can be made about the distance between the vertices of a triangle and the circumcenter?

Mathematics

1 answer:

slamgirl [31]3 years ago

7 0

Answer:

O the circumcenter is equidistant to all three vertices

Step-by-step explanation:

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Which recursive formula can be used to generate the sequence shown, where f(1) = 9.6 and n > 1? 9.6, 4.8, 2.4, 1.2, 0.6, ..

White raven [17]

Hello,

The formula is $f(n)= \dfrac{f(n-1)}{2}

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3 years ago

-4(x+1)+5=17 solve for x

ahrayia [7]

-4(x + 1) + 5 = 17

Distribute -4 inside the parentheses.

(-4x - 4) + 5 = 17

Combine like terms (-4 + 5).

-4x + 1 = 17

Subtract 1 from both sides.

-4x = 16

Divide both sides by -4.

x = -4

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3 years ago

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of the 480 canned foods collected, 45% of them were canned vegetables. How many canned vegetables did the school collect last mo

Alex787 [66]

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3 years ago

In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 g

KIM [24]

Answer:

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

All girls from the first group:

20 students, so $N = 20$

10 girls, so $k = 10$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163$

All girls from the second group:

20 students, so $N = 20$

5 girls, so $k = 5$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006$

All girls from the third group:

20 students, so $N = 20$

8 girls, so $k = 8$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036$

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

$P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}$

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

7 0

3 years ago

I need help with this question. It's about inequalities.

PtichkaEL [24]

N is greater than it equal to 3

So you put a shaded in for a three and draw a line to the right

I think :)

5 0

3 years ago