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MariettaO [177]
3 years ago
10

Sydney reads 50 pages in 40 minutes how many pages does she read per minute

Mathematics
2 answers:
Harman [31]3 years ago
5 0

"Per" in this context essentially means "divided by." To find pages divided by minutes, use the given numbers:

... (50 pages)/(40 minutes) = (50/40) pages/minute = 1.25 pages/minute

kirill115 [55]3 years ago
3 0

1.25 pages per minute.

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In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion
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Answer:

99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Step-by-step explanation:

We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.

Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.

Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;

                              P.Q. = \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n__1+_n__2-2

where, \bar X_1 = average gold thickness of control-immersion tip plating = 1.5 μm

\bar X_2 = average gold thickness of total immersion plating = 1.0 μm

s_1 = sample standard deviation of control-immersion tip plating = 0.25 μm

s_2 = sample standard deviation of total immersion plating = 0.15 μm

n_1 = sample of printed circuit edge connectors plated with control-immersion tip plating = 7

n_2 = sample of connectors plated with total immersion plating = 5

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }   =  \sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2}  }{7+5-2} }  = 0.216

<em>Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

So, 99% confidence interval for the difference between the mean population mean, (\mu_1-\mu_2) is ;

P(-3.169 < t_1_0 < 3.169) = 0.99  {As the critical value of t at 10 degree of

                                              freedom are -3.169 & 3.169 with P = 0.5%}  

P(-3.169 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } < 3.169) = 0.99

P( -3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } ) = 0.99

P( (\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } ) = 0.99

<u>99% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } , (\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } ]

= [ (1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5}  } } , (1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5}  } } ]

= [0.099 μm , 0.901 μm]

Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

6 0
3 years ago
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