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timofeeve [1]
1 year ago
13

The construction of copying ∠"QPR" is started below. The next step is to set the width of the compass to the length of ("AB" ).

How does this step ensure that the new angle will be congruent to the original angle?

Mathematics
1 answer:
Roman55 [17]1 year ago
4 0

Answer:

Constant radius, AB = TU

Step-by-step explanation:

Construction: Copying of <QPR, so that;

         <QPR = <TSU

The compass is placed in such a way that its two pointed legs are place at points A and B respectively to copy distance AB.

Now, place the compass at T and describe an arc to locate point U. Draw a straight line through S and U.

Therefore,

             <QPR = <TSU

The step ensures that the angles are congruent by constant radius, AB = TU.

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In triangle ABC, m∠BAC = 50°. If m∠ACB = 30°, then the triangle is triangle. If m∠ABC = 40°, then the triangle is triangle. If t
Agata [3.3K]
Acb is an OBTUSE triangle.
Abc is a RIGHT triangle.
Then AC= 4 or 6
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8 0
1 year ago
Read 2 more answers
CHEGG When testing the claim that bags of M&amp;M candies will have less than 3% broken pieces, which of the following would be
Vadim26 [7]

Answer:

H0: p ≥ 0.03 Ha: p < 0.03

Step-by-step explanation:

1) Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

2) Solution to the problem

On this case we want to test is p, where:

p represent the true proportion for the population of broken pieces

What we want to proof need's to be on the alternative hypothesis and the complement on the null hypothesis.

So the correct system of hypothesis for this case would be:

Null hypothesis: p\geq 0.03

Alternative hypothesis: p < 0.03

H0: p ≥ 0.03 Ha: p < 0.03

5 0
1 year ago
Marshall's mother buys 2 boxes of granola bars each week. Each box contains 8 granola bars. If she continues buying 2 boxes each
frutty [35]
2 x 8 =16
16 granola bars a week 
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7 0
1 year ago
Assume there are 365 days in a year.
MissTica

Answer:

1) The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.

Step-by-step explanation:

Given : Assume there are 365 days in a year.

To find : 1) What is the probability that ten students in a class have different birthdays?

2) What is the probability that among ten students in a class, at least two of them share a birthday?

Solution :

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

Total outcome = 365

1) Probability that ten students in a class have different birthdays is

The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...

\frac{364}{365}\times \frac{363}{365} \times \frac{362}{365} \times \frac{361}{365}\times\frac{360}{365} \times \frac{359}{365} \times \frac{358}{365} \times \frac{357}{365} \times\frac{356}{365}=0.883

The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday

P(2 born on same day) = 1- P( 2 not born on same day)

\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]

\text{P(2 born on same day) }=1-[\frac{364}{365}]

\text{P(2 born on same day) }=0.002

The probability that among ten students in a class, at least two of them share a birthday is 0.002.

6 0
2 years ago
Two production lines produce the same parts per week of which 100 are defective. Line 2 produces 2,000 parts per week of which 1
Vikentia [17]

Answer:

(a) 0.0833 or 8.33%

(b) 0.40 or 40%

Step-by-step explanation:

Parts line one (n1) = 1,000 parts

Defects line one (d1) = 100 parts

Parts line two (n2) = 2,000 parts

Defects line two (d2) = 150 parts

Total number of parts (n) = 3,000 parts

a. Probability of a randomly selected part being defective:

P(d) = \frac{d_1+d_2}{n_1+n_2}=\frac{100+150}{1,000+2,000}\\P(d) =0.0833=8.33\%

The probability is 0.0833 or 8.33%

b. Probability of a part being produced by line one, given that it is defective:

P(1|d)=\frac{P(1\cap d)}{P(1\cap d)+P(2\cap d)}\\P(1|d)=\frac{\frac{100}{3,000} }{\frac{100}{3,000}+\frac{150}{3,000}}\\P(1|d)=0.4 = 40\%

The probability is 0.40 or 40%.

5 0
1 year ago
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