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3 years ago

How is making a perpendicular bisector different to making an angle bisector

Mathematics

1 answer:

mr_godi [17]3 years ago

3 0

Answer and Step-by-step explanation:

When making a perpendicular line, you take 2 points on a straight line.

- You put the compass on the two points, making marks above and below to be able to draw the line.

When making an angle bisector line, you use the angle and 2 points to make a straight line that will bisect the angle.

- You put the compass on the angle, making an arc from the two lines that join to make the angle. Then, you put your compass on where the arc meets the lines, and make a mark, which you then draw a line through.

<u><em>#teamtrees #PAW (Plant and Water)</em></u>

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A variable whose value is less than zero is called ____________

pochemuha

It is called a negative number. For example, -4 -976 -23 -21

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4 years ago

Enter the ordered pair that is the solution to the system of equations graphed below.

kakasveta [241]

Answer:

Step-by-step explanation:

The solution is the coordinates of the point where the lines intersect: (3,5)

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3 years ago

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

frozen [14]

Note: Consider we need to find the vertices of the triangle A'B'C'

Given:

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).

To find:

The vertices of the triangle A'B'C'.

Solution:

If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then

$(x,y)\to (y,-x)$

Using this rule, we get

$A(-3,6)\to A'(6,3)$

$B(2,9)\to B'(9,-2)$

$C(1,1)\to C'(1,-1)$

Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).

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4 years ago

Can someone help me out please

Akimi4 [234]

Here are some of them

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3 years ago

Read 2 more answers

A very large batch of parts (assume a normal distrbution0 from a manufacturer has a mean weight = 43g and a standard deviation =

AVprozaik [17]

Answer:

5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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.

Percentage of parts with weights exceeding 45g?

1 subtracted by the pvalue of Z when X = 45. So

We have $\mu = 43, \sigma = 4$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 43}{4}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

1 - 0.6915 = 0.3075

If you slect 16 parts at random form that batch, what is the probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g?

This is P(X = 8) when n = 16, p = 0.3075. So

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

$P(X = 8) = C_{16,8}.(0.3075)^{8}.(0.6915)^{8} = 0.0544$

5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g

4 0

3 years ago