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2 years ago

What is the equation of the line that passes through the point (-4,8) and has a slope of -1/4

Mathematics

1 answer:

iragen [17]2 years ago

3 0

Answer:

Step-by-step explanation:

When a slope of a line and a point passing through it is given then we use slope - one point form to determine the equation of the line.

It is given by:

$(y - y_1) = m (x - x_1)$

where $m$ is the slope of the line and

$(x_1, y_1)$ is the point passing through it.

Here, $m = \frac{-1}{4}$ and $(x_1,y_1) = (-4,8)$

Substituting in the equation:

$y - 8 = \frac{-1}{4} (x + 4)$

$\implies 4y - 32 = -x - 4$

$\implies x + 4y - 28 = 0$

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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

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

2 years ago

Greta has watched 30 minutes of a movie. This is 20% of the entire movie. How long is the movie?

german

The answer is 180 minutes. because you multiply 30 times 0.20 and you get 6. Then you multiply 6 times 30 and get 180 minutes. And 180 minutes in hours is 3 hours.

8 0

2 years ago

Emile will use less than 1/2 cup of sugar for a recipe what fraction might emile use?

Mandarinka [93]

Answer:

1/4, 1/8, 1/5, *Actually there are a lot of possibilities.*

Step-by-step explanation:

7 0

1 year ago

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What kind of number is -2

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