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horsena [70]
3 years ago
11

for men age 18-24 in the HANES5 sample, the regression equation forpredicting height from weight ispredicted height=(0.0267 inch

es per pound) * (weight)+65.2inches(height is measured in inches and weight in pounds.) if someoneputs on 20 pounds, will he get taller by20 pounds* 0.0267 inches per pound ˜ 0.5 inches?
Mathematics
1 answer:
Harrizon [31]3 years ago
7 0

Answer:

When someone puts on 20 pounds then, the rate of change of height or the height increases by 0.534 inches.                  

Step-by-step explanation:

We are given the following equation:

Predicted height = (0.0267 inches per pound)(weight) + 65.2

Let y b the predicted height in inches and x be the weight in pounds.

y =65.2 + 0.0267x

Comparing to general form of equation:

y = mx +c

where m is the slope and c is the y-intercept.

m = 0.0267

Interpretation of slope:

Th slope tells us about the rate of change that is when the weight increases by 1 unit that is 1 pound then, the height increases by 0.0267 inches.

When someone puts on 20 pounds:

y(x+20)-y(x) =65.2 + 0.0267(x+20) - 65.2 - 0.0267x\\y(x+20)-y(x) = 0.534

Interpretation:

Thus, when someone puts on 20 pounds then, the rate of change of height or the height increases by 0.534 inches.

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1 year 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.

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Write a sequence with exactly 5 terms and follow the pattern rule:
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Answer:

The pattern. 1, 4, 9, 16, … has no common difference. An explicit pattern rule is a pattern rule that tells you how to get any term in the pattern without listing all the terms before it. For example, an explicit pattern rule for 5, 8, 11, 14, … uses the first term (5) and the common difference (3).

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Answer:

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mart [117]

Answer:

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Step-by-step explanation:

we know that

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so

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so

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