All categories

2 years ago

PLS HELP ME

Mathematics

1 answer:

eimsori [14]2 years ago

6 0

Answer:

5.5

Step-by-step explanation:

2,3,3,4,5,7,8,8,8,10,10,12

The mean of this data is 7.5. So what you are trying to figure out is the mean between 2 and 7. Since that is 3 and 4 it would be 3.5. So 3.5 is the Q1. Now you have to figure out the mean between 8 and 12. That is 9. This is Q3. You have to subtract 9-3.5 and that is the interquartile range. So 5.5 is the answer. I hope that made sense.

You might be interested in

Find the slope of the line whose equation is 0.05x - 0.03y = 9.

trapecia [35]

Using the slope-intercept form (y=mx+b) the slope is:
1.66666666

5 0

3 years ago

Read 2 more answers

8+345737377948585848484

iris [78.8K]

Answer:

3.4573738e+20

Step-by-step explanation:

6 0

3 years ago

What is 9.375 Rounded to the nearest hundredth

stiks02 [169]

Note the hundredth place value (underlined and bolded):

9.375

Look at the number to the right of the hundredth place value. It is a 5. Because you round up if the number is 5 or greater, you round up in this case (you round down if it is 4 or less).

9.375 rounded to the nearest hundredth is 9.38

3 0

3 years ago

Read 2 more answers

X-4 /16=9 1/3 A- -13 1/2 b- -5 1/6 c- 13 1/2 d- 5 1/6

4vir4ik [10]

Answer:

<h2>hope it helps you see the attachment for further information (๏_๏)(๏_๏)(๏_๏)</h2>

4 0

3 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

2 years ago