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3 years ago

How would I solve this?? Any help is much appreciated

Mathematics

1 answer:

kirill [66]3 years ago

4 0

Answer:

155 km

Step-by-step explanation:

As shown:

At Δ BDC which is a right triangle at ∠D

BC = 78 km

m∠CBD = 90° - 64° = 26°

cos ∠CBD = BD/BC

∴ BD = BC * cos ∠CBD = 78 * cos 26° = 70.106 km

Port C is at a distace to the east of A = AO + BD

= 85 + 70.106 = 155.106 km ≈ 155 km to the nearest km

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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

1. Compare 2.01 x 1015 and 2.8 x 1013. Which number is larger?

Leya [2.2K]

Answer:

If these are in exponent form -- the answer would be 2.01 x $10^{15}$ > 2.8 x 10.3

Step-by-step explanation:

Look at the sheet attached.

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3 years ago

Which statement best describes the faces that make up the total surface area of this composite solid? A triangular prism on top

Anestetic [448]

Answer:

The answer is B) 9 faces, 7 rectangles, and 2 triangles

Step-by-step explanation:

I took the test and got it correct

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3 years ago

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Given that |-6| > 5, which statement is not true?

azamat

-6 is greater than 5 is wrong

7 0

2 years ago

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A family read the odometer before starting a trip .It read 22900.6 and they know they will be driving 302.5 miles . What's will

AlexFokin [52]

Answer:

23,203.1

Step-by-step explanation:

Generally, an odometer measures distance traveled during the usage life cycle of a car. It is necessary to say the distance is measured cumulatively.

Now, to get the value the odometer will be reading after the trip, we need to add the present reading of the odometer to the new distance to be traveled.

Thus, the reading of the odometer after the trip will be 22900.6 + 302.5 = 23,203.1

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3 years ago