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bulgar [2K]
3 years ago
12

4.000 mm= how many cm? Check the comments who ever helps me thank you

Mathematics
2 answers:
Alexus [3.1K]3 years ago
7 0

Answer:0.4 cm

Step-by-step explanation:

Leya [2.2K]3 years ago
5 0

Answer:

0.4 cm

Step-by-step explanation:

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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.

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Find the midpoint of the segment connecting (4,7) to (-2,3).
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If you put 2 points on a graph paper and connect them, you can find the midpoint pretty easily
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Simplify 5(x - 6).<br><br> -x<br> 5x - 6<br> 5x - 30
Tamiku [17]
Hello! For this problem, we have to use the distributive property. So we do 5 * x and 5 * 6. When we do that, we get 5x - 30. The answer is C: 5x - 30.
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3 years ago
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Which of the following measurements is a little more than a yard? A. Centiliter B. 1/4 foot C. Centimeter D. Meter
BabaBlast [244]

D? Maybe I think I’m right hopefully!!

3 0
3 years ago
Help, please
stich3 [128]

The width of the rectangle is 6.82 meters, while the length of the rectangle is 8.184 meters.

<h3>What is a rectangle?</h3>

A parallelogram in which adjacent sides are perpendicular to each other is called a rectangle. A rectangle is always a parallelogram and a quadrilateral but the reverse statement may or may not be true.

Let the width of the rectangle be x.

Given that a rectangular room is 1.2 times as long as it is wide. Therefore, the length is 1.2x.

Perimeter = 2(x + 1.2x)

30 meters = 2(2.2x)

30 = 4.4x

x = 6.82

Hence, the width of the rectangle is 6.82 meters, while the length of the rectangle is 8.184 meters.

Learn more about Rectangle:

brainly.com/question/15019502

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5 0
1 year ago
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