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DENIUS [597]
3 years ago
13

I need help!! I'll report if wrong ​

Mathematics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

23

Step-by-step explanation:

2+2+3+4+5+6 = 23

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Can you find the perimeter of rectangle if you are given juat the measure of two sides?
Naddika [18.5K]
It depends, if the two sides given are adjacent to each other then yes. but you are given two opposite sides then no, because you don't know the length of one of the sides, thus not being able to calculate the perimeter
6 0
3 years ago
It costs 135.25 for 15 child tickets ti the museum.Adult tickets cost 12.95 each. Find the cost for 40 children to attend along
Masja [62]

Answer:the cost for 40 children to attend along with 5 adult tickets = $ 417.55

Step-by-step explanation:

It costs 135.25 for 15 child tickets to the museum. Let us determine the cost per child ticket to the museum.

cost per child ticket to the museum.

= total cost of 15 child tickets / number of child tickets.

= 132.25 / 15 = 8.82

Adult tickets cost 12.95 each.

The cost of 40 child tickets is

40 × 8.82 = $352.8

The cost of 5 adult tickets = 5 × 12.95

=$ 64.75

the cost for 40 children to attend along with 5 adult tickets would be

352.8 + 64.75 = $ 417.55

4 0
2 years ago
Round the number to the place of the underlined digit.<br><br> 4.327 (The underlined Digit is 3)
Genrish500 [490]

Answer:4.3

You round to the tenths place

5 0
2 years ago
Write the slope intercept form of the equation of the line through the given point with the given slope through (-2,5),slope-3/2
hjlf
Answer: y= -3/2x + 2
y= -3/2 + b
5 = -3/2 * -2 + b
5 = 3 + b
2 = b
6 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:

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
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