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sammy [17]
2 years ago
15

Describe at least two similarities between constructing a perpendicular line through a point on a line and constructing a perpen

dicular through a point off a line. (10 points)
Mathematics
1 answer:
Vitek1552 [10]2 years ago
3 0

The similarities between constructing a perpendicular line through a point on a line and constructing a perpendicular through a point off a line include:

  • Both methods involve making a 90-degree angle between two lines.
  • The methods determine a point equidistant from two equidistant points on the line.

<h3>What are perpendicular lines?</h3>

Perpendicular lines are defined as two lines that meet or intersect each other at right angles.

In this case, both methods involve making a 90-degree angle between two lines and the methods determine a point equidistant from two equidistant points on the line.

Learn more about perpendicular lines on:

brainly.com/question/7098341

#SPJ1

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How does this diagram help show that 2/7 = 8/28
Sveta_85 [38]
This diagram shows how 2/7 = 8/28 by showing that you can color in more squares or duplicate that amount of squares 4 times in order to get 8/28. (Since 2/7 x 4/4 = 8/28.)
5 0
3 years ago
Which is the quotient for 28/8?<br><br> A. 0.25<br><br> B. 0.35<br><br> C. 3.25<br><br> D. 3.5
Marysya12 [62]

Answer:

D. 3.5

Step-by-step explanation:

28 divided by 8 = 3.5

4 0
2 years ago
Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list
Radda [10]

Answer:

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: Any from 0 to 5.

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

Step-by-step explanation:

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.

In this question:

n = 5, p = 0.3, q = 1 - p = 0.7

So

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

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

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

8 0
3 years ago
Can someone help pls?!?!?
Reil [10]

to factor it the answer is (x+1)(x+1)

to simplify it the answer is x2+2x+1

idk if this helped or not but here you go

8 0
3 years ago
If you made $300 one week and had a total of $75 worth of deductions, how much did you actually take home?
UNO [17]

Answer:

$225

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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