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yuradex [85]
2 years ago
5

If Angle 6 is congruent to angle 10 and Angle 5 is congruent to angle 7, which describes all the lines that must be parallel? Li

nes r and s are crossed by lines t and u to form 16 angles. Clockwise from top left, at the intersection of r and t, the angles are 1, 2, 3, 4; at the intersection of s and t, 5, 6, 7, 8; at the intersection of u and s, 9, 10, 11, 12; at the intersection of u and r, 13, 14, 15, 16. Only lines r and s must be parallel. Only lines t and u must be parallel. Lines r and s and lines t and u must be parallel. Neither lines r and s nor lines t and u must be parallel.

Mathematics
2 answers:
mel-nik [20]2 years ago
8 0

Answer:

quick answer is the Third option-  Lines "r" and "s" and lines "t" and "u" must be parallel.

Nina [5.8K]2 years ago
7 0

Answer:

Third option: Lines "r" and "s" and lines "t" and "u" must be parallel.

Step-by-step explanation:

The missing figure is attached.

You need to remember that:

1- A Transversal is defined as a line that intersects two or more lines.

2- When a transversal cut two parallel lines, several angles are formed, which are grouped in pairs. Some of them are:

a. Vertical angles: are those pairs of angles that share the same vertex and are opposite to each other. These angles are congruent.

b. Corresponding angles: are those  pairs of non-adjacent angles located on the same side of the transversal and outside the parallel lines. They are congruent.

In this case, you can identify in the figure that:

 \angle 6 and \angle 10 are Corresponding angles.

\angle5 and \angle 7 are Vertical angles.

Therefore, based on the explained before, you can conclude that lines "r" and "s" and lines "t" and "u", must be parallel.

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A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr
erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034

The estimated standard error is:

\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

8 0
2 years ago
As part of a class project, a university student surveyed the students in the cafeteria lunch line to look for a relationship be
Vedmedyk [2.9K]

Answer:

0.63

Step-by-step explanation:

To get the relative frequency of students with red hair, we need to find the relative frequency of students with red hair from the students with gray eyes.

Since we already know that the total number of students with gray eyes is 35 and the number of red hair students with gray eye is 22

The formula we will use to calculate the relative frequency will be

The relative frequency of students with red hair = Total students of red hair with gray eyes divide by total number of gray eye students

22 ÷ 35 = 0.628  ≅ 0.63

6 1
2 years ago
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,
Gre4nikov [31]

Answer:

A.the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

\mathtt{H_o: \mu = 100}

\mathtt{H_1: \mu \neq 100}

A. If the acceptance region is defined as 98.5 <  \overline x >  101.5 , find the type I error probability \alpha .

Assuming the critical region lies within \overline x < 98.5 or \overline x > 101.5, for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is \mu = 100

∴

\mathtt{\alpha = P( type  \ 1  \ error ) = P( reject \  H_o)}

\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5  )}

when  \mu = 100

\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}  > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }

\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z  > \dfrac{1.5}{\dfrac{2}{3}}) }

\mathtt{\alpha = P ( Z  2.25) }

\mathtt{\alpha = P ( Z

From the standard normal distribution tables

\mathtt{\alpha = 0.0122+( 1-  0.9878) })

\mathtt{\alpha = 0.0122+( 0.0122) })

\mathbf{\alpha = 0.0244 }

Thus, the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis \mathtt{H_o}

Thus;

β = P( type II error) - P( fail to reject \mathtt{H_o} )

∴

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 103

\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }

\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 105

\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }

\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0
2 years ago
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dolphi86 [110]

Let the price of 1 shirt be represented by = s

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