DUDE I NEEED HELP FRIST ANSWER GET BRIANLY

PLEASE HELP!!! In the diagram below, if m1 = 79, find the measure of each numbered angle.

Reika [66]

2 - 101
3 - 101
4 - 79
5 - 79
6 - 101
lol hope this helps <33

3 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

a line is perpendicular to y=-4x-2 and intersects the point (0,9). what is the equation for this perpendicular line?

AysviL [449]

Answer:

y=(1/4)x + 9 OR y=x/4 +9

Step-by-step explanation:

Perpendicular lines have a slope/gradient of -1/n

Because the slope of this line is -4, the slope of a line perpendicular to it would be 1/4.

y=(1/4)x+b

(0,9) is the y-intercept, so b = 9

y=(1/4)x + 9 OR y=x/4 +9

3 0

4 years ago

20 POINTS! + BRAINLIEST!

g100num [7]

Answer:

= x^4 -x^3 -4x^2 -3

Step-by-step explanation:

f(x) = x^4 -x^2 +9

g(x) = x^3 +3x^2 +12

We want to find f-g

(f-g) (x) = x^4 -x^2 +9 - (x^3 +3x^2 +12)

Distribute the minus sign

(f-g) (x) = x^4 -x^2 +9 - x^3 -3x^2 -12

= x^4 -x^3 -4x^2 -3

3 0

3 years ago

Read 2 more answers

Given the diagram determine the height of the tree. round answer to the nearest tenth of a foot

Mice21 [21]

the answer should be 70
base times height divided by 2
equals the answer for triangles

4 0

4 years ago