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Sophie [7]
3 years ago
11

-3/4 + 1/5 find each sum with out using a number line

Mathematics
1 answer:
Maru [420]3 years ago
8 0
Might be terribly wrong since it's been a while that I've done fractions, but I got -11/20. Hope this helps!!!
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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
How to simplify 52 to 24
Mila [183]
Find the GCD (or HCF) of numerator and denominator
GCD of 52 and 24 is 4
Divide both the numerator and denominator by the GCD
52 ÷ 4
24 ÷ 4
Reduced fraction: 13/6
5 0
3 years ago
At a water park, 4 out of 20 tickets sold were child tickets. What percentage of the tickets were child tickets?
jeyben [28]

Answer:

20%

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
For Bill's birthday his mom is bringing donuts to school.
kipiarov [429]

36 donuts (2 1/2 dozen) = $8.00

So, you divide 8 by 36 and you get about 22¢ for each donut. Then, you do .22 x 12 which equals $2.64.

One dozen donuts = $2.64

3 0
3 years ago
An isosceles triangle has 2 equivalent sides. The perimeter of an isosceles triangle is 84 centimeters. The measure of one of th
Pani-rosa [81]

Well, to find your solution to this problem, I would subtract 84 from 32 = 52. I would divide the number by 2, which equals 26. So, side lengths are 32, 26, and 26. 

Or, another one is that 2 sides have the length of 32, so 32+32=64. I would subtract 84 from 64 and get 20. So, side lengths are 32, 32, and 20.

3 0
3 years ago
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