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Zarrin [17]
4 years ago
13

Please help me I’m so lost

Mathematics
2 answers:
kogti [31]4 years ago
8 0

Answer:

331.4

Step-by-step explanation:

Side length (a)

8.284

yd

Perimeter

66.27

yd

Area

331.4

yd²

Longest diagonal (l)

21.65

yd

Medium diagonal (m)

20

yd

Shortest diagonal (s)

15.307

yd

Circumcircle radius (R)

10.824

yd

Incircle radius (r)

10

yd

Ugo [173]4 years ago
8 0
The answer is 66.27 yes
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The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the f
geniusboy [140]

Answer:

The answer is "266.5".

Step-by-step explanation:

Given value:

115, 370, 250, 93, 540, 225, 177, 425, 318, 182, 275, and\  228.

find:

Sample mean=?

=\frac{\text{total dental expenses}}{number \ of \ employees}

=\frac{115+370+250+93+540+ 225+ 177+ 425+ 318+ 182+ 275+228}{12}\\\\=\frac{3,198}{12}\\\\=266.5\\\\

7 0
3 years ago
Two basketball teams, the Tigers and the Bears, played several games. The Tigers won the first game by two points. The total num
aleksandrvk [35]

Answer: Tigers = 46     Bears = 44

<u>Step-by-step explanation:</u>

Let x represent the numbers of points scored by the Bears.

then x + 2 is the number of points scored by the Tigers.

Bears + Tigers = Total

  x      +  x + 2  =   90

            2x + 2 =   90

            2x       =    88

              x       =    44

Bears (x) = 44

Tigers (x + 2) = 44 + 2 = 46

5 0
3 years ago
What is the missing step in the proof?
Cloud [144]
I think is a because if you see and do it you could see
8 0
3 years ago
Least to greatest .5 1/5 .35 12/25 4/5
N76 [4]
1/5, .35, 12/25, .50, 4/5
3 0
3 years ago
Problem 5. Buses arrive at 116th and Broadway at the times of a Poisson arrival process with intensity λ arrivals per hour. Thes
Shtirlitz [24]

Answer:

Step-by-step explanation:

The problem states that there are only two types of busses - M104 and M6 with probable occurence of 0.6 and 0.4 respectively.

If the average number of busses arriving per hour is λ, the average number of M6 busses per hour is  0.4λ

Now consider a set of 3 M6 busses as an event. The average number of such events per hour will be

μ = 0.4λ / 3

The expected number of hours for the event "THIRD M6 arrives", let's say X is

E[X] = 1 / μ ( exponential distribution) = 3 / 0.4λ

= 7.5 / λ

The variance of event X is =

Var[x] = \frac{1}{U^2} = \frac{56.25}{\lambda ^2}

6 0
3 years ago
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