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seropon [69]
3 years ago
6

The Saxena family plans to install a light to illuminate part of their rectangular yard. Nikki and Dylan each proposed a differe

nt spot to place the light. The proposed placements and the lit area that each produces are shown below. Nikki's proposed placemat has a triangular light area with a base of 60 feet and height of 38 feet. Dylan's proposed placemat has a base of 60 feet and a height of 38 feet. How do Nikki's and Dylan's proposals compare? Check all that apply.

Mathematics
1 answer:
monitta3 years ago
3 0

Options:

a. Nikki's proposed placement will light a greater area than Dylan's placement.

b.  Dylan's proposed placement will light a greater area than Nikki's placement.

c.  Both proposed placements will light the same sized area.

d. Nikki's proposed placement will light more than half the yard.

e. Dylan's proposed placement will light more than half the yard.

f.  Dylan's proposed placement will light exactly half of the yard.

g. Nikki's proposed placement will light less than half of the yard.

Answer:

C) Both proposed placements will light the same sized area.

F) Dylan's proposed placement will light exactly half of the yard.

Step-by-step explanation:

The area of a triangle is (base x height) / 2, and both lights illuminate the same base and height = (60 x 38) / 2 = 1,140 sq ft

Both Dylan's and Nikki's proposed placement will lit exactly half of the yard. The yard's total area = 60 x 38 = 2,280 sq ft, which is twice the area lit by the lights.

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Elenna [48]
<h2>Answer: </h2>

<u>In a right triangle, Relation of base and hypotenuse is cosine</u>.

cos40° = 10/h

0.76604 = 10 × h

0.76604h = 10

h = 10/0.76604 = 13.05units.

7 0
2 years ago
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
2 years ago
An electronics hobbyist has three electronic parts cabinets with two drawers each.
Andrews [41]

Answer:

a) 0.5 = 50% probability that an NPN transistor will be selected.

b) 0.3333 = 33.33% probability that it came from the cabinet that contains both types

c) 66.67% probability that it comes from the cabinet that contains only NPN transistors

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

a) What is the probability that an NPN transistor will be selected?

1/3 probability that the first cabinet is chosen. This cabinet has two transistors, both of which are NPN, so 100% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, both of which are PNP, so 0% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, one of which is NPN, so 50% probability of selecting a NPN transistor.

So

p = \frac{1}{3}*1 + \frac{1}{3}*0 + \frac{1}{3}*0.5 = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = 0.5

0.5 = 50% probability that an NPN transistor will be selected.

b) Given that the hobbyist selects an NPN transistor, what is the probability that it came from the cabinet that contains both types?

Here we use the conditional probability formula.

Event A: NPN transistor

Event B: From the third cabinet.

50% probability that an NPN transistor will be selected, so P(A) = 0.5.

1/6 probability that it is from the third cabinet and NPN, so P(A \cap B) = \frac{1}{6}

The desired probability is:

P(B|A) = \frac{\frac{1}{6}}{0.5} = 0.3333

0.3333 = 33.33% probability that it came from the cabinet that contains both types.

c) Given that an NPN transistor is selected what is the probability that it comes from the cabinet that contains only NPN transistors?

Either it comes from the cabinet with only NPN transistors, or it comes from the cabinet with both types of transistors. The sum of the probabilities of these outcomes is 100%. So

x + 33.33 = 100

x = 66.67

66.67% probability that it comes from the cabinet that contains only NPN transistors

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Answer:

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Step-by-step explanation:

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