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Dvinal [7]
2 years ago
7

In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. A long-distance telephone c

ompany wants to perform a hypothesis test to determine whether the mean duration of long-distance phone calls has changed from the 1990 mean of 7.2 minutes. H0: μ < 7.2 minutes Ha: μ > 7.2 minutes H0: μ = 7.2 minutes Ha: μ ≠ 7.2 minutes H0: μ = 7.2 minutes Ha: μ ≤ 7.2 minutes H0: μ ≠ 7.2 minutes Ha: μ = 7.2 minutes
Mathematics
1 answer:
scoundrel [369]2 years ago
4 0

Answer:

We know that In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. And we want to test if the mean duration of long-distance phone calls has changed from the 1990 mean of 7.2 minutes (alternative hypothesis) and the complement rule would represent the null hypothesis.

The correct system of hypothesis are:

Null hypothesis: \mu =7.2

Alternative hypothesis: \mu \neq 7.2

So then the best option for this case would be:

H0: μ = 7.2 minutes Ha: μ ≠ 7.2 minutes

Step-by-step explanation:

We know that In 1990, the mean duration of long-distance telephone calls originating in one town was 7.2 minutes. And we want to test if the mean duration of long-distance phone calls has changed from the 1990 mean of 7.2 minutes (alternative hypothesis) and the complement rule would represent the null hypothesis.

The correct system of hypothesis are:

Null hypothesis: \mu =7.2

Alternative hypothesis: \mu \neq 7.2

So then the best option for this case would be:

H0: μ = 7.2 minutes Ha: μ ≠ 7.2 minutes

And in order to test the hypothesis we can use a one sample t test or z test depending if we know the population deviation or not

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Anton will be constructing a segment bisector with a compass and straightedge, while Maxim will be constructing an angle bisecto
Genrish500 [490]

The similarities are;

  • Compass and a straight edge required for both construction
  • Both construction includes a line drawn from the intersection of arcs to bisect a segment or an angle
  • The bases for the construction of both bisector are the ends of segment and the angle to be bisected
  • The width of the compass when drawing intersecting arcs, is more than half the width of the segment or angle being bisected

The differences are;

  • Two points of intersection of arcs are used in the segment bisector while only one is requited in an angle bisector
  • The bisecting line crosses the segment in a segment bisector, while it stops at the vertex of the angle being bisected in an angle bisector

The sources of the above equations are as follows;

The steps to construct a segment bisector are;

  • Place the needle of the compass at one of the ends of the line segment to be bisected
  • Widen the compass so as to extend more than half of the length of the segment to be bisected
  • Draw two arcs, one above, and the other below the line
  • Place the compass needle at the other end and with the same compass width draw arcs that intersects with the arcs drawn in the above step
  • Draw a line segment by placing the ruler on the points of intersection of the arcs above and below the line

The steps to construct an angle bisector are;

  • With the compass needle at the vertex, open the pencil end such that arcs can be drawn on the rays (lines) forming the angle
  • Draw an arc on both lines forming the angle
  • Place the compass needle at one of the intersection points and draw an arc in between the lines forming the angle
  • Repeat the above step with the same compass width from the other intersection point with the rays forming the angle
  • Join the point of intersection of the two arcs to the vertex of the angle to bisect the angle

Therefore, we have;

The similarities are;

  • A compass and a straight edge can be used for both construction
  • A straight line is drawn from the point of intersection of arcs to bisect the segment or the angle
  • The arcs are drawn from the ends of the segment or angle to be bisected
  • The width of the compass is more than half the width of the line or angle when drawing the arcs

The differences are;

  • In a segment bisector, the intersection point is above and below the line, while in an angle bisector only one pair of arcs are drawn to intersect above the line
  • The bisecting line passes through the segment being bisected, while the line stops at the vertex in an angle bisector

Learn more about the construction of segment and angle bisectors here;

brainly.com/question/17335869

brainly.com/question/12028523

7 0
1 year ago
Match the following items.
Margarita [4]

Answer:

1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40

Step-by-step explanation:

Given:

\angle DBC=40°

From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.

m\textrm{ arc CD}=2\times \angle DBC\\m\textrm{ arc CD}=2\times 40=80

Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,

m\angle BCD=90

From the Δ BCD, which is a right angled triangle,

m\angle DBC+m\angle BDC=90\textrm{ (right angled triangle)}\\40+m\angle BDC=90\\m\angle BDC=90-40=50

Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.

Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles m\angle DBC\textrm{ and }m\angle BDC and EF is a tangent to the circle at point C.

Therefore, m\angle DCF=m\angle DBC=40\\m\angle ECB=m\angle BDC=50

Again, using the same theorem as above,

m\angle DCF=50\\\therefore m\textrm{ arc BC}=2\times m\angle DCF=2\times 50=100

Hence, all the angles are as follows:

1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40

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If two triangles are congruent, then they have equal corresponding angles and also the sides.
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