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Effectus [21]
3 years ago
14

Calculate the distance between the points K = (1, -1) and P=(9.-6) in the coordinate plane.

Mathematics
1 answer:
GarryVolchara [31]3 years ago
6 0

Answer:

√89

Step-by-step explanation:

√(x2 - x1)² + (y2 - y1)²

√(9 - 1)² + [-6 - (-1)]²

√(8)² + (-5)²

√64 + 25

√89

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Which of the following, is not a solution for the inequalities is represented by the graph shown?
stealth61 [152]

Answer:

(0,1)

Step-by-step explanation:

This is because (0,1) is right on the line, while the other choices are inside of the blue shaded area.

If this answer is correct, please make me Brainliest!

3 0
3 years ago
HEPL MEE, I NEED EXTREME HELP!
natta225 [31]

Answer:

A= ( 5 x 3 ) x [(9-5)x3 divided by 2]

15+6

=21

P= 5+5+3+9

=22

6 0
3 years ago
Melinda earns 5 minutes to play video games for every hour she spends on homework.
Law Incorporation [45]
<h2>Answer:</h2>

15 minutes

<h2>Step-by-step explanation:</h2>

For every hour of homework she gets 5 minutes of video games.

She spend 3 hours of homework.

We multiply 3*5

Answer is 15

8 0
3 years ago
Read 2 more answers
And those... sorryyyyyy!:(
sasho [114]

Answer:

F: 18

Step-by-step explanation:

find total amount of points

43 + 7 = 50

subtract half of 50 to find the midpoint

43 - 25

18

L = 18

3 0
3 years ago
Read 2 more answers
Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
Zarrin [17]

Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population​ mean is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

             \mu = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.878 < t_1_8 < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 22.4 -2.878 \times {\frac{3.8}{\sqrt{19} } , 22.4 +2.878 \times {\frac{3.8}{\sqrt{19} } ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

6 0
3 years ago
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