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ELEN [110]
3 years ago
7

Another way to write 1/2

Mathematics
2 answers:
Pie3 years ago
5 0
Another way to write 1/2 is .5
earnstyle [38]3 years ago
3 0
½=0.5.................
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A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed
AURORKA [14]

Answer:

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.

Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;

                     P.Q.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean speed for the 25-mil film = 1.15

\bar X_1 = sample mean speed for the 20-mil film = 1.06

s_1 = sample standard deviation for the 25-mil film = 0.11

s_2 = sample standard deviation for the 20-mil film = 0.09

n_1 = sample of 25-mil film = 8

n_2 = sample of 20-mil film = 8

\mu_1 = population mean speed for the 25-mil film

\mu_2 = population mean speed for the 20-mil film

Also,  s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} } = \sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} } = 0.1005

<em>Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.</em>

<u>So, 98% confidence interval for the difference in population means, (</u>\mu_1-\mu_2<u>) is;</u>

P(-2.624 < t_1_4 < 2.624) = 0.98  {As the critical value of t at 14 degrees of

                                             freedom are -2.624 & 2.624 with P = 1%}  

P(-2.624 < \frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } < 2.624) = 0.98

P( -2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } < 2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } <  ) = 0.98

P( (\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } ) = 0.98

<u>98% confidence interval for</u> (\mu_1-\mu_2) = [ (\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } ]

= [ (1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } } , (1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } } ]

 = [-0.042, 0.222]

Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.

7 0
2 years ago
Please help mee!!!<br><br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B6%7D%7B-2%7D%3D-1" id="TexFormula1" title="\frac{x+6}
Misha Larkins [42]
Multiply both sides with -2
X + 6 = 2
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3 0
2 years ago
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3 years ago
There was d dollars in the savings account when mr Duane withdrew $45 from the account. Then me Duane deposited and additional $
Basile [38]
What are you asking ???
Write it in an equation ???
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6 0
3 years ago
Which of the following statements correctly illustrate the addition property of equality?
liubo4ka [24]

Answer:

x - y + z = 8

2x +3y -z= -2

3x - 2y -9z = 9

Step-by-step explanation:

[1 - 1  1  8]

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[3 -2 -9  9]

4 0
2 years ago
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