LMN and QRS are similar. Find the value of X.

Select "Yes" or "No" to indicate whether the ordered pair is on the graph of the function f(x)=−16x+1 .

alexandr402 [8]

To check if the pair is on the graph, plug the x value into the x in the equation, and solve. If the solution to the equation matches the y value from your ordered pair, it’s on the graph, so mark “yes”. If the solution isn’t the same, it’s not on the graph, so mark “no”.

8 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

2 years ago

(a) An angle measures 140 . What is the measure of its supplement? (b) An angle measures 36 . What is the measure of its complem

Lesechka [4]

A: 40 because supplements add up to 180
B: 54 because complements add up to 90

6 0

3 years ago

Susan needs 228 ounces of fruit juice for a party. She is choosing between two brands of juice. She finds the unit rate of each

Sergio [31]

Answer:

"Brand A costs approximately $0.21 per ounce, and Brand B costs approximately $0.18 per ounce."

In order to find the rounded cost per one item (in this case, per one ounce) Susan needs to divide the total price between the number of units and after that, round the result obtained up to the nearest cent.

Therefore:

Brand A

2.55$/12 ounces= 0.2125 $/ounce

As the third decimal digit, 2, is closer to 0 than to 9, then we maintain the second decimal digit as 1.

The price per unit of brand A after rounding it up is 0.21 $ per ounce

Brand B

1.45$/8 ounces= 0.1812

As the third decimal digit, 1, is closer to 0 than to 9, then we maintain the second decimal digit as 8.

The price per unit of brand B after rounding it up is 0.18 $ per ounce

4 0

3 years ago

What is the estimate of 92 times 68

solong [7]

92 can be rounded to 90
68 can be rounded to 70

90 x 70 = 6300

8 0

3 years ago

Read 2 more answers

LMN and QRS are similar. Find the value of X.​

LMN and QRS are similar. Find the value of X.