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Rainbow [258]
2 years ago
11

Find the circumference of a circle that has a radius of 36 inches

Mathematics
1 answer:
SVEN [57.7K]2 years ago
7 0
I think that the answer is 226.194 inches
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Anyone?????????????????
k0ka [10]
X < 6

circle not highlighted
6 0
3 years ago
Read 2 more answers
Combine the like terms<br> 2g + 15 + 3g + g + g- 18
ZanzabumX [31]

Answer:

7g - 3

Step-by-step explanation:

2g + 15 + 3g + g + g - 18

2g + 3g + g + g + 15 - 18

7g - 3

6 0
3 years ago
I need help with this problem,
emmasim [6.3K]

Answer:

  JL = 78

Step-by-step explanation:

The shorter segment is a midline, so is half the length of the longer one.

  2(5x-16) = 4x +34

  5x -16 = 2x +17 . . . . . divide by 2

  3x = 33 . . . . . . . . . . add 16-2x

  x = 11 . . . . . . . . . . divide by 3

Then segment JL is ...

  JL = 4x +34 = 4(11) +34 = 44+34

  JL = 78

4 0
2 years ago
A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs
Makovka662 [10]

Answer:

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be​ 95% confident that the mean​ cost, μ​, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the mean​cost, μ​, of all recent weddings in this country = (22,550.95, 30,226.40)

.The​ 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be​ 95% confident that the mean​ cost, μ​,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this​ interval, but we can be​ 95% confident that it does.

Hope this Helps!!!

4 0
3 years ago
An experiment consists of drawing 1 card from a standard​ 52-card deck. let e be the event that card drawn is a 33. find​ p(e).
Dafna1 [17]

1 out of 52. So 1/52 equals some decimal. Plug in calc.

4 0
3 years ago
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