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Setler [38]
2 years ago
12

I need to evaluate the limit and I also need to show the work for it. Please help me out if you can.

Mathematics
1 answer:
True [87]2 years ago
6 0

Answer:

-6

Step-by-step explanation:

You can solve all math problems via  Photo math application, you can download it from the store for free, it is easy to write the question via the camera and quickly display the final result with the solution steps.

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12. Last year, a toy company produced 115,000 board games. This was a 4% decrease in production over the previous year. Which eq
Travka [436]

Answer:

B

Step-by-step explanation

115000 = 0.96p

115000 = p - 0.04p

I think It's B if the = sign in the beginning is a - sign

7 0
2 years ago
Write an equation in slope intercept form for the line with slope 1/2 and y-intercept -2
forsale [732]
Y=1/2x + -2 i think
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3 years ago
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Is there a x=-2 in these equations
dem82 [27]

plug in the x for all of those to find y


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3 years ago
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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:

A)

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

10.7

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

C)

8.3

,10.3

,10.5

,10.6

,10.7

,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
When fractions share the same number on the bottom what is it called?
monitta
That means they have a common denominator. Hope this helps!
5 0
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