What is a answer that has been rounded calked

Mathematics

1 answer:

Tomtit [17]3 years ago

5 0

Answer:

guess

Step-by-step explanation:

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15. In the rhombus shown below, mz1 = 160°. What are the measures of 22 2 points

egoroff_w [7]

Answers:

angle2 = 160 degrees

angle3 = 10 degrees

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Explanation:

The two triangles are congruent. We can prove this using SSS. The upper triangle is a mirrored copy of the lower triangle.

Since angle1 = 160 degrees, this means angle2 = 160 also. This is because the corresponding pieces of congruent triangles match up. Or you could use the second part of the hint which says "Opposite angles in a rhombus are congruent".

Let x be the measure of angle 3. For either isosceles triangle, it has congruent base angles measured x degrees. They add to the 160 degrees mentioned earlier to get a total of 180.

160+x+x = 180

2x+160 = 180

2x = 180-160

2x = 20

x = 20/2

x = 10

Angle 3 is 10 degrees.

6 0

2 years ago

There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one

diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

$P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02$

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01$

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05$

So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:

P(A)=1/3

P(B)=1/3

P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C$