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wariber [46]
2 years ago
8

What types of concurrent constructions are needed to find the orthocenter of a triangle? A. intersection of the lines drawn perp

endicular to each side of the triangle through its midpoint B. intersection of the lines drawn to bisect each vertex of the triangle C. intersection of the lines drawn from each vertex of the triangle and perpendicular to its opposite side D. intersection of the lines drawn to the midpoint of each side of the triangle to its opposite vertex
Mathematics
1 answer:
finlep [7]2 years ago
4 0

Answer: Hello! The answer to your question is B, the intersection of the lines drawn to bisect each vertex of the triangle. Hope this helped! Please pick my answer as the Brainliest!

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There are 20 students in a math class who have brown hair. This represents 80 percent of the students in the class. Which equati
Reptile [31]

Answer:

The answer is C

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
If I chose a number uniformly from the integers from 1 to 25, calculate the conditional probability that the number is a multipl
topjm [15]

Answer:

<h2>1/7</h2>

Step-by-step explanation:

If I choose a number from the integers 1 to 25, the total number of integers I can pick is the total outcome which is 25. n(U) = 25

Let the probability that the number chosen at random is a multiple of  6 be P(A) and the probability that the number chosen at random is is larger than 18 be P(B)

P(A) = P(multiple of 6)

P(B) = P(number larger than 18)

A = {6, 12, 18, 24}

B = {19, 20, 21, 22, 23, 24, 25}

The conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is expressed as P(A|B).

P(A|B) = P(A∩B)/P(B)

Since probability = expected outcome/total outcome

A∩B = {24}

n(A∩B) = 1

P(A∩B) = n(A∩B)/n(U)

P(A∩B) = 1/25

Given B = {19, 20, 21, 22, 23, 24, 25}.

n(B) = 7

p(B) = n(B)/n(U)

p(B) = 7/25

Since P(A|B) = P(A∩B)/P(B)

P(A|B) = (1/25)/(7/24)

P(A|B) = 1/25*25/7

P(A|B) = 1/7

<em></em>

<em>Hence the conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is 1/7</em>

7 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
s2008m [1.1K]

Answer: 912

===================================

Work Shown:

The starting term is a1 = 3. The common difference is d = 5 (since we add 5 to each term to get the next term). The nth term formula is

an = a1+d(n-1)

an = 3+5(n-1)

an = 3+5n-5

an = 5n-2

Plug n = 19 into the formula to find the 19th term

an = 5n-2

a19 = 5*19-2

a19 = 95-2

a19 = 93

Add the first and nineteenth terms (a1 = 3 and a19 = 93) to get a1+a19 = 3+93 = 96

Multiply this by n/2 = 19/2 = 9.5 to get the final answer

96*9.5 = 912

I used the formula

Sn = (n/2)*(a1 + an)

where you add the first term (a1) to the nth term (an), then multiply by n/2

-----------------

As a check, here are the 19 terms listed out and added up. We get 912 like expected.

3+8+13+18   +23+28+33+38    +43+48+53+58    +63+68+73+78   +83+88+93 = 912

There are 19 values being added up in that equation above. I used spaces to help group the values (groups of four; except the last group which is 3 values) so it's a bit more readable.

5 0
3 years ago
Read 2 more answers
If 1once of toppings costs. .50 how much would 11onces be
dybincka [34]

Answer:

If one ounce of toppings cost $0.50, then you would need to take $0.50*11

.5*11= 5.5

11 ounces would be $5.50

hope this helps ;)

6 0
3 years ago
If k is a positive integer and n = k(k + 7), is n divisible by 6 ? (1) k is odd. (2) When k is divided by 3, the remainder is 2.
olganol [36]

Answer:

1.) Yes

2.) Yes

Step-by-step explanation:

Given that

n = k(k + 7)

If k is a positive integer and n = k(k + 7), is n divisible by 6 ?

(1) k is odd. Yes.

Let assume that k = 3

Then, n = 3(3 + 7)

n = 3 × 10

n = 30.

30 is divisible by 6.

(2) When k is divided by 3, the remainder is 2. That is,

Let k = 5

Then,

5/3 = 1 remainder 2

Substitute k into the equation

n = k(k + 7)

n = 5(5 + 7)

n = 5 × 12

n = 60

And 60 is divisible by 6.

Therefore, the answer to both questions is Yes.

5 0
3 years ago
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