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Ivanshal [37]
1 year ago
7

I need help on this question

Mathematics
1 answer:
vodka [1.7K]1 year ago
6 0

Answer

Arc EF = 52°

Arc HD = 142°

Angle HGF = 128°

Explanation

To solve for the unknown angles, we need to first solve for x

To do that, we need to first note that the sum of angles on a straight line is 180°

So,

Angle HCG + Angle HCD = 180° (Sum of angles on a straight line)

Angle HCG = 2x

Angle HCD = 6x + 28°

Angle HCG + Angle HCD = 180°

2x + 6x + 28° = 180°

8x + 28° = 180°

8x = 180° - 28°

8x = 152°

Divide both sides by 8

(8x/8) = (152°/8)

x = 19°

Angle HCG = 2x = 2 (19°) = 38°

Angle HCD = 6x + 28° = 6(19°) + 28° = 142°

So, we can solve for the rest now

Arc EF = Angle ECF

= 90° - Angle ECD

Angle ECD = Angle HCG = 38° (Vertically opposite angles are equal)

Arc EF = Angle ECF

= 90° - Angle ECD

= 90° - 38°

= 52°

Arc HD = Angle HCD = 142°

Angle HGF = Angle HCG + Angle GCF = 38° + 90° = 128°

Hope this Helps!!!

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Answer:

a) Probability that vehicle is Light Truck is 0.69

b) Probability that vehicle is imported car is 0.08 or 8%.

Step-by-step explanation:

Part a)

Event that vehicle is a car is given by A. Since, the vehicles can only be classified as cars or light trucks, the event that a vehicle will be light truck will be compliment of A.

i.e. Event that vehicle is a Light Truck = A^{c}

It is given that in recent year 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks.

So, from here we can say that, if a vehicle is randomly selected from all the vehicles the probability that it would be a light truck will be:

P(Vehicle will be Light Truck) = P(A^{c}) = 69% = 0.69

Part b)

Event that vehicle is imported is given by B. We need to find the probability the a randomly chosen vehicle is an imported car i.e. we have to find probability of occurrence of events A and B together, which will be denoted as: P(A and B)

Since, P(A^{c}) = 69% = 0.69

P(A) = 1 - P(A^{c}) = 1 - 0.69 = 0.31

It is also given that 78% of vehicles were domestic. This means, the percentage/probability of imported vehicles is:

P(B) = 1 - 0.78 = 0.22

55% of vehicles were domestic light trucks. This can be expressed as:

P(A^{c}\&B^{c}), the compliment of this event will give us P(A or B).

i.e.

P(A or B) = 1 - P(A^{c} \&B^{c}) = 1 - 0.55 = 0.45

According to the addition rule of probabilities:

P(A or B) = P(A) + P(B) - P(A and B)

Substituting the calculated values gives us:

0.45 = 0.31 + 0.22 - P(A and B)

P(A and B) = 0.08

This means, the probability that vehicle is imported car is 0.08 or 8%.

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Hello,
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