Need help with number 13 plz :D

Order each set of measures from greatest to least 1) 1.2 cm, 0.6 in, 0.031 m, 0.1 ft 2) 2 lb, 891 g, 1 kg, 0.02 T 3) 1 1/4 c, 0.

labwork [276]

Step-by-step explanation:

1) 1.2 cm, 0.6 in, 0.031 m, 0.1 ft

The relationship between the units is given as;

m > ft > in > cm

Upon converting the rest units to the lowest;

0.6 in = 1.524 cm

0.031 m = 3.1 cm

0.1 ft = 3.048 cm

The order is given as;

0.031 m > 0.1 ft > 0.6 in > 1.2 cm

2) 2 lb, 891 g, 1 kg, 0.02 T

2 lb = 907.185 g

1 kg = 1000 g

0.02 T = 20000 g

The order is given as;

0.02 T > 1 kg > 2 lb > 891 g

3) 1 1/4 c, 0.4 L, 950 mL, 0.7 gal

0.4 L = 400 mL

0.7 gal = 2649.79 mL

11/4 c = 650.6177 mL

The order is given as;

0.7 gal > 11/4 c > 950ml > 0.4 L

4) 4.5 ft, 48 in., 1.3 m, 120 cm

The relationship between the units is given as;

m > ft > in > cm

Upon converting the rest units to the lowest;

4.5 ft = 137.16 cm

48 in = 121.92 cm

1.3 m = 130 cm

The order is given as;

4.5 ft > 1.3 m > 48 in > 120 cm

3 0

3 years ago

Which value represents the slope of the line that has an x-intercept at (-4,0) and a y-intercept at (0, -12) in the coordinate p

Lesechka [4]

To find the gradient you do change in y/change in x so...

-12 - 0
————- = -12/4 = -3
0 - - 4

Hope this helps! Any questions let me know :)

6 0

3 years ago

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

3.(4+5) is how much more than 3.4+5?

kifflom [539]

Answer:10 more

Step-by-step explanation:

distributive property so 3*4+3*5 so 12 an 15 or 27. 3*4+5 is 12+5 so 17 27-17 is 10

5 0

3 years ago

Read 2 more answers

Which statement is logically equivalent to the following conditional statement?

anastassius [24]

Hi there! Your answer would be:

D, If it is an octagon, then it does not have exactly five sides.

I also know because I took this exam and it was correct.

Hope that helps!

6 0

2 years ago

Read 2 more answers