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2 years ago

Nayan plays different games in the play ground from 6.15 to 7.00 in the morning and

Mathematics

2 answers:

Likurg_2 [28]2 years ago

6 0

Answer:

nayan plays for 45 min

Step-by-step explanation:

count up

jek_recluse [69]2 years ago

6 0

Answer:

The answer is either 45 minutes or 90 minutes / 1 and a half hours

Step-by-step explanation:

You either add the two times together, as he plays for 45 minutes at one time and another 45 minutes at a later time. I'm not sure if the question is asking for the sum or how much time he plays for as an average because they are both 45 minutes.

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How do you do this question?

tatyana61 [14]

Answer:

Step-by-step explanation:

The increments are 0.4, and the width of the interval is 10, so the number of increments (the number of times you need to use Euler's method) is:

10 / 0.4 = 25

8 0

2 years ago

A survey was conducted that asked 1002 people how many books they had read in the past year. results indicated that x overbar eq

Shalnov [3]

The confidence interval would be (10.44, 12.16). This means that if we take repeated samples, the true mean lies in 90% of these intervals.

To find the confidence interval, we use:
$\overline{x} \pm z*(\frac{\sigma}{\sqrt{n}})$

We first find the z-value associated with this. To do this:
Convert 90% to a decimal: 90% = 90/100 = 0.9
Subtract from 1: 1-0.9 = 0.1
Divide by 2: 0.1/2 = 0.05
Subtract from 1: 1-0.05 = 0.95

Using a z-table (http://www.z-table.com) we see that this is directly between two z-scores, 1.64 and 1.65; we will use 1.645:

$11.3 \pm 1.645*(\frac{16.6}{\sqrt{1002}})
\\
\\11.3 \pm 0.86
\\(11.3-0.86, 11.3+0.86)
\\(10.44, 12.16)$

3 0

3 years ago

1.)34°C=__°F<br>2.)94.6°F=__°C<br>3.)80°F=__°C<br>4.)42°C=__°F<br>5.)18°C=__°F

Leto [7]

Answer:

Answer:

Step-by-step explanation:

1.)93.2F

2.)34.78°C

3.)26.67°C

4.)107.6°F

5.)64.4°F

I hope it's helpful!

6 0

2 years ago

A classroom has 4 new boxes of chalk and 6 individual pieces of chalk in use. How many total pieces of chalk are in the classroo

AveGali [126]

The total number of pieces of chalk in the classroom will be the number of pieces in the 4 boxes in addition to the six individual pieces.

The average box of chalk holds 12 pieces of chalk
This means that 4 boxes have 12*4 = 48 pieces

Total number of pieces = 48 + 6 = 54 pieces of chalk

8 0

2 years ago

For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need to calculate these statistics;

umka2103 [35]

Answer:

a) The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Step-by-step explanation:

For any distribution,

The median is the variable at the middle when all the variables in the dsitribution are arranged in ascending or descending order.

The median is the (n+1)/2 th variabke.

where n = size of the distribution or number of variables. For these questions, the sample size is 5, hence, the median will always be the 3rd variable when the variables are arranged in ascending or descending order.

The IQR, known as the inter quartile range is a measure of dispersion for the distribution. Although, it isn't as effective as other measures of dispersion such as the standard deviation because the IQR unlike the the standard deviation isn't responsive to changes in the variables, especially ones at the end of the distribution.

The IQR is simply given mathematically as the third quartile minus the first quartile.

IQR = (Third quartile) - (First quartile)

Third quartile is the 3(n+1)/4 th variable. For a sample of n=5, the third quartile is the 4.5th variable, that is the average of the 4th and 5th variable.

First quartile is the (n+1)/4 th variable. For a sample of n=5, the first quartile is the 1.5th variable, that is the average of the 1st and 2nd variable.

Taking the questions one at a time

a) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,6,7,20

Median = 3rd variable = 6

Third quartile = (7+20)/2 = 13.5

First quartile = (3+5)/2 = 4

IQR = 13.5 - 4 = 9.5

The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) (1) 1,2,3,4,5

Median = 3rd variable = 3

Third quartile = (4+5)/2 = 4.5

First quartile = (1+2)/2 = 1.5

IQR = 4.5 - 1.5 = 3

(2) 6,7,8,9,10

Median = 3rd variable = 8

Third quartile = (9+10)/2 = 9.5

First quartile = (6+7)/2 = 6.5

IQR = 9.5 - 6.5 = 3

The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,7,8,9

Median = 3rd variable = 7

Third quartile = (8+9)/2 = 8.5

First quartile = (3+5)/2 = 4

IQR = 8.5 - 4 = 4.5

The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) (1) 0,10,50,60,100

Median = 3rd variable = 50

Third quartile = (60+100)/2 = 80

First quartile = (0+10)/2 = 5

IQR = 80 - 5 = 75

(2) 0,100,500,600,1000

Median = 3rd variable = 500

Third quartile = (600+1000)/2 =800

First quartile = (0+100)/2 = 50

IQR = 800 - 50 = 750

The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Hope this Helps!!!

5 0

3 years ago