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tatiyna
3 years ago
11

What is the first three terms of 8-2

Mathematics
1 answer:
zheka24 [161]3 years ago
4 0
The answer to this question is what the person said above!! Have a nice day!!
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A cone has a volume of 120 pi cubic feet. If the radius is 3 feet, what
sergey [27]

Answer:

Step-by-step explanation:

V= 2pi^h

120=2(3,14) h

120/6,28= 19 feet

6 0
3 years ago
One inch is 2.5cm how many inches are there in 24.4cm? (2 d.p)
olganol [36]


one inch---> 24.4 cm

inches in 24.4cm=> 24.4/2.5

=> 9.76cm (2 decimal places)

6 0
3 years ago
Read 2 more answers
What is the value of x?<br><br> Enter your answer, as a decimal, in the box.<br><br><br> m
UkoKoshka [18]
X / 96.6 =  (80.5 - 35) / 80.5  ( because the small and large triangles are similar,  because of the parallel lines)

x / 96.6 =  45.5 / 80.5

x = 96.6 * 45.5  / 80.5 =  54.6 m (answer)
7 0
4 years ago
Read 2 more answers
A space is totally disconnected if its connected spaces are one-point-sets.Show that a finite Hausdorff space is totally disconn
marysya [2.9K]

Step-by-step explanation:

If X is a finite Hausdorff space then every two points of X can be separated by open neighborhoods. Say the points of X are x_1, x_2, ..., x_n. So there are disjoint open neighborhoods U_{12} and U_2, of x_1 and x_2 respectively (that's the definition of Hausdorff space). There are also open disjoint neighborhoods U_{13} and U_3 of x_1 and x_3 respectively, and disjoint open neighborhoods U_{14} and U_4 of x_1 and x_4, and so on, all the way to disjoint open neighborhoods U_{1n}, and U_n of x_1 and x_n respectively. So U=U_2 \cup U_3 \cup ... \cup U_n has every element of X in it, except for x_1. Since U is union of open sets, it is open, and so U^c, which is the singleton \{ x_1\}, is closed. Therefore every singleton is closed.

Now, remember finite union of closed sets is closed, so \{ x_2\} \cup \{ x_3\} \cup ... \cup \{ x_n\} is closed, and so its complemented, which is \{ x_1\} is open. Therefore every singleton is also open.

That means any two points of X belong to different connected components (since we can express X as the union of the open sets \{ x_1\} \cup \{ x_2,...,x_n\}, so that x_1 is in a different connected component than x_2,...,x_n, and same could be done with any x_i), and so each point is in its own connected component. And so the space is totally disconnected.

4 0
4 years ago
1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
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