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Elis [28]
3 years ago
14

Gwen, Tristan, and Keith like to work out at the gym. They each have an app on their phone that estimates the calories they burn

. By the time Tristan and Keith started exercising, Gwen had already burned 100 calories. Gwen burns 10 calories every minute, Tristan burns 125 every 10 minutes, and Keith burns 300 calories every 30 minutes. Let x represent the number of minutes since Tristan and Keith started exercising and y represent the number of app-estimated calories burned.

Mathematics
2 answers:
Setler [38]3 years ago
7 0

Answer: ?

Step-by-step explanation:

aleksandrvk [35]3 years ago
6 0

Answer:

Part A:

Gwen has already burned 100 calories when Tristan and Keith started, and he continues to burn 10 calories per minute. The equation for Gwen is

y = 100 + 10x

Part B:

Tristan is burning 125 calories every 10 minutes. His burning calories rate is 12.5 calories per minute. The equation for Tristan is

y = 12.5x

Part C:

Keith is burning 300 calories in 30 minutes. His rate per minute is 10 calories per minute. The equation for Keith is

y = 10x

Part D:

Just follow the instructions and do the thing for the app. You can then answer the questions it is very easy.

Part E:

Do the same for E after you do Part D.

Step-by-step explanation:

Hope it helps!

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Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

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

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

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