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aniked [119]
2 years ago
12

6. a) The mid point of a line segment is (1,-1) whose one end is (-1, -3). Find the other end. ​

Mathematics
1 answer:
Anna [14]2 years ago
5 0

Answer:

Step-by-step explanation:

mid point = [(a1 + a2)/2 , (b1 + b2)]

(a1,b1) = first point

(a2,b2) = end point

Using all x coordinates

(a1 + a2)/2 = 1

(-1 + a2)/2 = 1

-1 + a2 = 2

a2 = 3

Using all y coordinates

(b1 + b2)/2 = -1

(-3 + b2)/2 = -1

-3 + b2 = -2

b2 = -1

thus end coordinate = (3, -1)

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I hope this helps you



regular hexagon =6 equilateral triangle



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Area=600
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1/4 x + 1/6 x = x-7.
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Answer:

x=12

Step-by-step explanation:

Simplify both sides by adding like terms

5/12x=x-7

subtract x from both sides

-7/12x=-7

multiply both sides by 12/-7

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3 years ago
Solve for x 73 =(9/10x)+19
Yuliya22 [10]
73=(\frac{9}{10}x)+19

To solve x:

1. Substract 19 in both sides of the equation:

\begin{gathered} 73-19=(\frac{9}{10}x)+19-19 \\  \\ 54=\frac{9}{10}x \end{gathered}

2. Multiply both sides of the ewuation by 10/9:

\begin{gathered} 54\cdot\frac{10}{9}=\frac{9}{10}x\cdot\frac{10}{9} \\  \\ \frac{540}{9}=x \\  \\ 60=x \end{gathered}<h2>Then, the value of x is 60 (x=60)</h2>
3 0
10 months ago
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jarptica [38.1K]

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6 0
3 years ago
​41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the
lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least​ six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which 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)!}

And p is the probability of X happening.

​41% of U.S. adults have very little confidence in newspapers.

This means that p = 0.41

You randomly select 10 U.S. adults.

This means that n = 10

(a) exactly​ five

This is P(X = 5). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

(b) at least​ six

This is:

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209

P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480

P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125

P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019

P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001

Then

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least​ six.

(c) less than four.

This is:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051

P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355

P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111

P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058

So

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0
3 years ago
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