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mihalych1998 [28]
3 years ago
5

Find y. A. √2/2 B. 4 C. √6/2 D. √2

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

Step-by-step explanation:

the answer is option D

WHEN YOU SUBSTITUTE THE VALUES YOU WILL GET IT............

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❗️❗️❗️Find the length of side x in simplest radical form with a rational denominator.❗️❗️❗️
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Answer:

3

Step-by-step explanation:

In a 30-60-90 triangle, the ratio of the shorter side to the longer side length is 1:\sqrt{3}. Therefore:

x=\sqrt{3} \cdot \sqrt{3}= \\\\\sqrt{3}^2= \\\\3

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Solve each inequality then graph the solution
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Answer:

a is x>3

b is _> x/4 +12

c is x>-3

Step-by-step explanation: hope this helps

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2 years ago
Samantha plays a trivia game. She earns 3 points for each
Rudiy27

この質問は緒と面白いね

11x3=33

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2 years ago
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8 0
2 years ago
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round y
kupik [55]

Answer:

75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.

Step-by-step explanation:

The question is missing. It is as follows:

Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69  104  125  129  60  64

Assume that the population of x values has an approximately normal distribution.

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

75% Confidence Interval can be calculated using M±ME where

  • M is the sample mean weight of the wild mountain lions (\frac{69 +104 +125 +129+60 +64}{6} =91.8)
  • ME is the margin of error of the mean

And margin of error (ME) of the mean can be calculated using the formula

ME=\frac{t*s}{\sqrt{N} } where

  • t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
  • s is the standard deviation of the sample(31.4)
  • N is the sample size (6)

Thus, ME=\frac{1.30*31.4}{\sqrt{6} } ≈16.66

Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5

3 0
3 years ago
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