Solve the equation by completing the square. Round to the nearest hundredth if necessary. x2

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

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.

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

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

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

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

= [-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

3 years ago

Does anybody know it part 2

makvit [3.9K]

Answer:

x = 33

Step-by-step explanation:

(x + 13) + (4x + 2) = 180

x + 13 + 4x + 2 = 180

5x + 15 = 180

5x = 180 - 15

5x = 165

x = 165/5

x = 33

Check:

(x + 13) + (4x + 2) = 180

(33 + 13) + 4(33) + 2 = 180

46 + 132 + 2 = 180

46 + 134 = 180

4 0

3 years ago

X(x+1)(x+3)(x+5) - x(x+3)(x+5)

mash [69]

Answer:

x^4 + 8x^3 + 15x^2

Step-by-step explanation:

8 0

3 years ago

Hamburger is on sale for 20% off the normal $3.70 price. How much can you save per pound by buying the meat on sale? Explain you

Nitella [24]

Answer:

$0.75

Step-by-step explanation:

Given

Normal price - $3.70

If Hamburger is on sale for 20% off the normal price

The amount that can be saved will be 20% of $3.70

Amount that can be saved = 20/100 * 3.70

Amount that can be saved = 1/5 * 3.70

Amount that can be saved = 3.7/5

Amount that can be saved = 0.75

Hence the Amount that can be saved on $3.70 is $0.75

3 0

3 years ago

In choosing this year 327 voted musical and 115 voted drama. How many more students voted musical than drama

Lapatulllka [165]

Answer:

212 more student voted musical than drama .

Step-by-step explanation:

As given

In choosing this year 327 voted musical and 115 voted drama.

More students voted musical than drama = Total students voted for musical - Students voted for drama .

Putting values in the above

More students voted musical than drama = 327- 115

= 212

Therefore 212 more student voted musical than drama .

5 0

3 years ago

Solve the equation by completing the square. Round to the nearest hundredth if necessary. x2 – x = 25