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3 years ago

Help me out please!!!~

Mathematics

2 answers:

Ipatiy [6.2K]3 years ago

7 0

I’m pretty sure it’s the 3rd one

kicyunya [14]3 years ago

3 0

Answer:

y = - 3/10x + 1/2

Step-by-step explanation:

It's the 2nd one because it's <u>rise</u>

run

Since the slope is negative, you're going down 3, right 10. and the y -intercept AKA the starting point, is 1/2 because it's on the y line & well, that's where it starts.

Hope this helps you!! :)

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Find the gradient of the line segment between the points (2,-3) and (-3,-8)

Anvisha [2.4K]

Answer:

Step-by-step explanation:

gradient=

$\frac{ - 8 - ( - 3)}{ - 3 - 2} = \frac{ - 8 + 3}{ - 5} = \frac{ - 5}{ - 5} = 1$

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3 years ago

PLEASE HELP!! THANK YOU <3

Natalija [7]

Answer:

what did you need to know :(

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3 years ago

Erin received the following scores on her last four spelling tests. 2125, 82%, 0.8, 1720 2125, 82%, 0.8, 1720 Which of these is

AveGali [126]

the highest score is obviously going to be 2125 because that is 100%

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2 years ago

2 MATH K.

Ann [662]

Answer:

Step-by-step explanation:

Given the addition problem:

58 + 36

We can form a table :

5 _______ 8

3 _______ 6

9 4

Hence, (58 + 36) = 94

4 0

2 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