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3 years ago

9

a fruit company delivers its fruit in two types of boxes large and small a delivery of five large boxes and two small boxes have

a total weight of 88kg delivery of three large boxes in 8 small boxes has a total of 97 kg how much does each type of box weigh

1 answer:

hodyreva [135]3 years ago

8 0

Answer:

Large boxes:19 kg

Small boxes:5 kg

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Graph a triangle (LMN) and rotate it 180° around the origin to create triangle L′M′N′.

andreev551 [17]

When a triangle is rotated, it must be rotated through a center of rotation.

Assume the coordinate of LMN is

$\mathbf{L = (1,1)}$

$\mathbf{M = (1,7)}$

$\mathbf{N = (3,6)}$

The rule of 180-degree rotation is:

$\mathbf{(x,y) \to (-x,-y)}$

So, we have:

$\mathbf{L' = (-1,-1)}$

$\mathbf{M' = (-1,-7)}$

$\mathbf{N' = (-3,-6)}$

See attachment for the graphs of LMN and L'M'N

<u>(a) Describe the transformation</u>

When triangle LMN is rotated 180 degrees across the origin, the x and y coordinates of LMN are negated to form triangle L'M'N'

<u>(b) Lines LL' and MM'</u>

The distance between point L and the center of origin is the same as the distance between point M' and the center of origin

Similarly, the distance between point L and the center of origin is the same as the distance between point M' and the center of origin

<u>(c) Line NN'</u>

Line NN' will assume the same characteristics as lines LL' and MM'

Read more about transformations at:

brainly.com/question/11707700

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2 years ago

The following are all angle measures (in degrees, rounded to the nearest tenth) whose tangent is 2.62.62, point, 6. Which is the

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Answer:

I'm not fully sure but id say the principle value would be choice C

Step-by-step explanation:

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2 years ago

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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

<em>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.</em>

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

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

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

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

= [-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.

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2 years ago

Brent would like to start a small landscape business this summer. He has a lawn mower, but needs a truck to carry the lawn mower

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I would buy the 2000 Toyota Tundra. Yes it is the most expensive, but it is the newest so it should require less expense money for problems, it is a manual transmission 6 cylinder so it should get better gas mileage, and it has fewer miles. The rankings of these trucks are not provided, however a Toyota tundra is considered to be a workhorse in the truck. If this is an opinion question, that's my choice.

Step-by-step explanation:

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