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Licemer1 [7]
2 years ago
8

The value of 3 in 920.003

Mathematics
2 answers:
pashok25 [27]2 years ago
8 0
Thousandths this is all to it
ArbitrLikvidat [17]2 years ago
5 0
The value of the 3 is 0.003, 3 thousandths
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The table gives predictions of the average costs of a gasoline car after the year 2030. Which graph best represents the relation
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It's the b graph

Step-by-step explanation:

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An elf ate 15 of your muffins. that was 5/8 of all of them. how many are left?
zavuch27 [327]

Answer: 9

Step-by-step explanation:

An elf ate 15 of your muffins and that was 5/8 of all of them. To get the number of muffins left goes thus:

We can first calculate the total number of muffins the person had. Let the total number of muffins be y. That means the elf ate 5/8 of y.

5/8 of y = 15

5/8 × y = 15

0.625 × y = 15

0.625y = 15

Divide both side by 0.625

0.625y/0.625 = 15/0.625

y = 24

The total amount of muffins is 24. Since the elf has eaten 15, the amount left will be: 24-15 = 9

6 0
2 years ago
which two points should the line of best fit go through to best represent the data in the scatter plot (1,16) and (2,15) (1,16)
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. (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
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