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algol13
3 years ago
13

1) The scatter plot below shows the distance Destiny traveled on her cycling trip for various lengths of time during the trip. W

hat equation could describe the line of best fit for the scatter plot?
2) Estimate an equation for the line of best fit for the following scatter plot for the post high school education and salary


please show your work

Mathematics
1 answer:
prisoha [69]3 years ago
4 0
To determine the equation of the best fit line, the graphical approach can be done. A line is used in between the given points. Then, the slope and the y-intercept of the line is determined to give us the equation. The equations of two scatter plots are
y = 0.22 x + 0.46
y = 8x + 8.25<span />
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Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,
REY [17]

Answer:

a) 217

b) 1351

c) 5403

Step-by-step explanation:

Given that:

confidence interval (c) = 0.95

\alpha =1-0.95=0.05\\\frac{\alpha }{2} =\frac{0.05}{2}=0.025

The Z score of \frac{\alpha }{2} is from the z table is given as:

Z_{\frac{\alpha }{2} }=Z_{0.025}=1.96

Range = $45000 - $30000 = $15000

The standard deviation (σ) is given as:

\sigma=\frac{Range}{4} =\frac{15000}{4}=3750

Sample size (n) is given as:

 n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2

a) E = $500

n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2= (\frac{1.96*3750}{500} )^2 ≈ 217

b) n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2= (\frac{1.96*3750}{200} )^2 ≈ 1351

c) n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2= (\frac{1.96*3750}{100} )^2 ≈ 5403

7 0
3 years ago
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-2x+5-40-9x - 13 help me please​
Naddika [18.5K]

Answer:

-48/11

Step-by-step explanation:

-2x+5-40-9x-13=0

-11x=40-5+13

-11x=48

x=-48/11

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4 years ago
Write the equation 12 − y = 2x − 5 in slope-intercept form. What are the slope and y-intercept?
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Slope: -2 
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7 0
3 years ago
Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0
3 years ago
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