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qwelly [4]
2 years ago
8

Mark uses an app that shows him how many kilometers he has run to prepare for a marathon. The app said he ran 9.654 kilometers.

He wants to post online how many miles he ran. Mark ran ___ miles.
Mathematics
1 answer:
Viefleur [7K]2 years ago
4 0

Answer:

5.9987174899

Step-by-step explanation:

9.654km / 1.609344

= 5.9987174899mi

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Simora [160]

2x-2y-6x-3-2y+6=(2x-6x)+(-2y-2y)+(-3+6)\\\\=-4x-4y+3\\\\3x+4y+2y-4+5y-10=(3x)+(4y+2y+5y)+(-4-10)\\\\=3x+11y-14\\\\3x-6y-6+2x+6y+2=(3x+2x)+(-6y+6y)+(-6+2)\\\\=5x-4\\\\8x+2y+3x-9y+3x^2-2y=(3x^2)+(8x+3x)+(2y-9y-2y)\\\\=3x^2+11x-9y

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1 cm^3 is equal to 1 mL so the answer would probably be 210 mL

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2 years ago
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Assuming that the population is normally​ distributed, construct a 9090​% confidence interval for the population mean for each o
jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

 

Set B:

Mean: 4.5

Standard dev: 2.45

 

% =  90 

Set A:

Standard Error, SE = s/ √n =    1.9/√8 = 0.67  

Degrees of freedom = n - 1 =   8 -1 =  7   

t- score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

 

Set B:

Standard Error, SE = s/ √n =    2.45/√8 = 0.87  

Degrees of freedom = n - 1 =   8 -1 = 7   

t- Score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

 

<span>We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.</span>

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  4. Similarly, in the fourth graph you can draw vertical lines that cross the graph twice
  5. The fifth graph is a function, because every vertical line crosses the graph once
  6. The last graph is a function, although discontinuous, for the same reason.
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