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3 years ago

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DescriptionIn mathematics, the Poincaré conjecture is a theorem about the characterization of the 3-sphere, which is the hypersp

here that bounds the unit ball in four-dimensional space. The conjecture states: Every simply connected, closed 3-manifold is homeomorphic to the 3-sphere.
Am I right

1 answer:

marishachu [46]3 years ago

6 0

I believe you are right

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Distance on graph

Mnenie [13.5K]

The graph can be divided into two shapes. A rectangle and a right triangle.

The distance that the rover will cover if it completes one circuit can be computed using the Pythagorean theorem and adding the sides of the rectangle.

Rectangle:
Width = 4 - 2 = 2 meters
Length = 11 - 2 = 9 meters

Triangle = 16 - 2 = 14 ; 14 - 2 = 12 meters (this is the short leg)
long leg of the triangle = length of the rectangle = 9 meters.

12² + 9² = c²
144 + 81 = c²
225 = c²
√225 = √c²
15 = c

Point A to B = 4 - 2 = 2 meters
Point B to C = 15 meters
Point C to D = 16 - 2 = 14 meters
Point D to A = 11 - 2 = 9 meters

Total distance traveled = 2 + 15 + 14 + 9 = 40 meters.

6 0

2 years ago

Are all irrational numbers real numbers. If so , why

Elena L [17]

Yes, all numbers are real numbers because a number is a number, and nothing can change that. I hope this helps you!

5 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

<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.

7 0

2 years ago

plz help whats the answer to this question !!! calculate the mass of a gold bar that is 18.00cm long, 9.21cm wide, and 4.45cm hi

erik [133]

We have to find the mass of the gold bar.

We have gold bar in the shape of a rectangular prism.

The length, width, and the height of the gold bar is 18.00 centimeters, 9.21 centimeters, and 4.45 centimeters respectively.

First of all we will find the volume of the gold bar which is given by the volume of rectangular prism:

Volume of the gold bar $= length \times width\times height$

Plugging the values in the equation we get,

Volume of the gold bar $=18.00 \times 9.21 \times 4.45= 737.721cm^3$

Now that we have the volume we can find the mass by using the formula,

$Mass= density \times volume$

The density of the gold is 19.32 grams per cubic centimeter. Plugging in the values of density and volume we get:

$Mass = 19.32\times 737.721=14252.769$ grams

So, the mass of the gold bar is 14252.769 grams

7 0

3 years ago

Josiah and his friends are going to the movies. Each ticket costs $10, and popcorn is $5 a bag. There is a $3 service fee for th

Free_Kalibri [48]

Make sure you check this because i do make mistakes.

10 x 4 = 40

75 - 40 = 35

So josiah has 35 dollars to spend for things other than tickets.

35 - 3 = 32 (for the service fee)

So josiah has 32 dollars to spend on popcorn

32/5 = 6.4 but you cant have half a bag of popcorn so

6 bags

8 0

2 years ago