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anastassius [24]
2 years ago
11

Think carefully about the angle relationship the following represents.

Mathematics
2 answers:
Igoryamba2 years ago
8 0

#1

  • Linear pair

\\ \rm\Rrightarrow 2x+48=180

\\ \rm\Rrightarrow x+24=90

\\ \rm\Rrightarrow x=66

#2

  • Complementary angles

\\ \rm\Rrightarrow 4x+7+35=90

\\ \rm\Rrightarrow 4x+42=90

\\ \rm\Rrightarrow 4x=48

\\ \rm\Rrightarrow x=12

  • 4x+7=4(12)+7=55°

#3

\\ \rm\Rrightarrow 3x+54=180

\\ \rm\Rrightarrow 3x=126

\\ \rm\Rrightarrow x=42

lisov135 [29]2 years ago
4 0

Answer:

See below ↓↓↓

Step-by-step explanation:

<u>Subquestion #1</u>

  • These angles form a linear pair
  • They are supplementary, and add up to 180°
  • 2x + 48 = 180
  • 2x = 132
  • <u>x = 66</u>

<u></u>

<u>Subquestion #2</u>

  • These angles are complementary, and add up to 90°
  • 35 + 4x + 7 = 90
  • 4x + 42 = 90
  • 4x = 48
  • <u>x = 12</u>

<u></u>

<u>Subquestion #3</u>

  • As in #1, they are a linear pair, and are supplementary
  • 3x + 54 = 180
  • 3x = 126
  • <u>x = 42</u>
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Dawn and abed each roll a standard die obtaining a random number from 1 to 6. if the probability that dawns number is larger tha
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a+b = 12

Step-by-step explanation:

There are 36 possible outcomes for rolling two dice. Let A be the number rolled by Abed and D be the number rolled by Dawn. The sample space, in the format [A,D], of all of the possibilities in which Dawn's number is larger is:

[5,6]

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[3.4] [3,5] [3.6]

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3 years ago
The mean percentange of a population of people eating out at least once a week is 57℅
Sidana [21]

Answer:

<u>The correct answer is B. between 56.45% and 57.55% </u>

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

Source: brainly.com/question/1068489

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week  = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

Standard deviation of the sample = 3.5/6.32

Standard deviation of the sample = 0.55

Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean  + 1 time the standard deviation of the sample. Thus:

57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1  * 0.55 = 57 + 0.55 = 57.55

<u>The correct answer is B. between 56.45% and 57.55% </u>

7 0
3 years ago
Match the reasons with the statements in the proof.
natta225 [31]
<span>1. m∠6=m∠8, b||c
Given

2. m∠7=m∠8
</span><span>If lines ||, corresponding angles =.
</span><span>
3. m∠6=m∠7
</span>Substitution 
<span>
4. a||b </span>
If alternate interior angles equal, then lines ||. 
6 0
3 years ago
Read 2 more answers
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