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vladimir1956 [14]

2 years ago

5

I need help sorry for bothering everyone. Two supporting reasons are missing from the proof. Complete the proof by dragging and

dropping the appropriate reasons into each of the empty boxes. Given: m∥nm∠2=60° Prove: m∠7=120

2 answers:

attashe74 [19]2 years ago

6 0

Since ∠2=60° = ∠8 so it will also be 60 °
the sum of the both are 60°+∠7=180°
∠7= 180°- 60° = 120°

inna [77]2 years ago

3 0

Answer:

Given:

$m \parallel n\\m\angle 2 = 60\°$

Prove:

$m\angle 7 = 120\°$

1. $m \parallel n\\m\angle 2 = 60\°$ , by Given.

2. $m\angle 2 = m\angle 8$ , by Alternate Exterior Angles Theorem.

3. $m\angle 8= 60\°$ , by substitution.

4. $m\angle 8 + m\angle 7 = 180\°$ , by Straight Angle Definition and Supplementary Angle Theorem.

5. $60\° +m\angle 7 = 180\°\\m\angle 7= 180\° - 60\°\\m\angle 7= 120\°$ , by substitution.

$\therefore m\angle 7 = 120\°$

It's important to remember that Alternate Exterior Angles are those which are placed outside the parallels and at different sides of the transversal, and these angles are always congruent.

Additionally, the definition of a straight angle is that this angle is equal to 180°, and supplementary angles theorem states that adjacent angles that sum 180° are supplementary one each other.

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The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and

Firdavs [7]

Answer:

1a

$P(39 < X < 48 ) = 0.8767$

1b

95% of all sample means will fall between $40.1 < \mu < 49.9$

1c

$\= x = 41. 795$

2

$n = 25$

Step-by-step explanation:

From the question we are told that

The mean is $n = 45$

The population standard deviation is $\sigma = 10$

The sample size is n = 16

Generally the standard error of the mean is mathematically represented as

$\sigma_{x} = \frac{ \sigma}{\sqrt{n} }$

=> $\sigma_{x} = \frac{ 10 }{\sqrt{16 } }$

=> $\sigma_{x} = 2.5$

Generally the probability that the sample mean will be between 39 and 48 minutes is

$P(39 < X < 48 ) = P( \frac{ 39 - 45}{ 2.5} < \frac{X - \mu }{\sigma } < \frac{ 48 - 45}{ 2.5} )$

=> $P(39 < X < 48 ) = P(-2.4 < Z< 1.2 )$

=> $P(39 < X < 48 ) = P( Z< 1.2 ) - P(Z < -2.4)$

From the z table the area under the normal curve to the left corresponding to 1.2 and -2.4 is

=> $P( Z< 1.2 ) = 0.88493$

and

$P( Z< - 2.4 ) = 0.0081975$

So

$P(39 < X < 48 ) = 0.88493 -0.0081975$

=> $P(39 < X < 48 ) = 0.8767$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * 2.5$

=> $E =4.9$

Generally the 95% of all sample means will fall between

$\mu -E < and \mu +E$

=> $45 -4.9\ and \ 45 + 4.9$

Generally the value which 90% of sample means is greater than is mathematically represented

$P( \= X > \= x ) = 0.90$

=> $P( \= X > \= x ) = P( \frac{\= X - \mu }{ \sigma_x} > \frac{\= x -45 }{ 2.5} ) = 0.90$

=> $P( \= X > \= x ) = P( Z > z ) = 0.90$

Generally from the z-table the critical value of 0.90 is

$z = -1.282$

$\frac{\= x -45 }{ 2.5} = -1.282$

=> $\= x = 41. 795$

Considering question 2

Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

$s = \frac{1}{5} \sigma$

Generally the standard deviation of the sample mean is mathematically represented as

$s = \frac{\sigma }{ \sqrt{n} }$

=> $\frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }$

=> $n = 5^2$

=> $n = 25$

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