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gtnhenbr [62]
3 years ago
11

When is the measure of a base angle

Mathematics
1 answer:
Vlada [557]3 years ago
8 0
Elaborate this question
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Please help!! will give brainliest ASAP
soldier1979 [14.2K]

Answer:

c = 36.06

Step-by-step explanation:

a² + b² = c²

20² + 30² = c²

400 + 900 = c²

c² = 1300

\sqrt{c^2} =\sqrt{1300}

c = 36.06

8 0
2 years ago
Simplify and find value. a 3x^3−x^3+5x^3−9x^3 if x=−1
balandron [24]

Answer:

I think it is 2

Sorry if I got it wrong

4 0
2 years ago
Read 2 more answers
Kray sells personalized photo frames. She charges $12 for the frame plus $0.25 for each letter the customer wants to engraved on
VARVARA [1.3K]

Answer:

c=0.25n+12

Step-by-step explanation:

6 0
3 years ago
. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0
2 years ago
If <br><img src="https://tex.z-dn.net/?f=%20%7B3%7D%5E%7B%20-%20x%7D%20%20%3D%20y" id="TexFormula1" title=" {3}^{ - x} = y" alt
o-na [289]
3^3x+1 is the answer if you are simplifying
7 0
2 years ago
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