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Elodia [21]
3 years ago
12

How do you calculate elapsed time?

Mathematics
2 answers:
Valentin [98]3 years ago
4 0
First tell me what elapsed time is
algol [13]3 years ago
3 0
You take the time the event eneded and subtract it from the time the event started. Say a concert starts at 9:30 and ends at 11:00, 11:00-9:30= 1:30 the total time elapsed is an hour and a half(1:30).
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Four people are carpeting a house. Heidi carpets 16 of the house, Lisa carpets 16% of the house, Martina carpets 0.2 of the hous
Alborosie

Answer:

Heidi, Lisa, Jaydie, and Martina

Step-by-step explanation:

16%, 16%, 18%, 20%

5 0
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Pls help me I only have 2 minutes left winner gets brainliest
devlian [24]

Answer:

i think it was about 5000.

Step-by-step explanation:

7 0
3 years ago
A class has fewer than 30 students exactly 2/3of the student's in the class own a football exactly 7/8 of the students in the cl
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This number would be 24. So 21 own football boots, 16 own a football and there are 24 children in the class.  not good at math so yeah
5 0
3 years ago
2 Construct a rational function that will help solve the problem. Then, use a calculator to answer the question.
vaieri [72.5K]

Answer:

<u><em>Dimension of box:-</em></u>

Side of square base = 10 in

Height of box = 5 in

Minimum Surface area, S = 300 in²

Step-by-step explanation:

An open box with a square base is to have a volume of 500 cubic inches.

Let side of the base be x and height of the box is y

Volume of box = area of base × height

                 500=x^2y

Therefore, y=\dfrac{500}{x^2}

It is open box. The surface area of box, S .

S=x^2+4xy

Put  y=\dfrac{500}{x^2}

S(x)=x^2+\dfrac{2000}{x}

This would be rational function of surface area.

For maximum/minimum to differentiate S(x)

S'(x)=2x-\dfrac{2000}{x^2}

For critical point, S'(x)=0

2x-\dfrac{2000}{x^2}=0

x^3=1000

x=10

Put x = 10 into y=\dfrac{500}{x^2}

y = 5

Double derivative of S(x)

S''(x)=2+\dfrac{4000}{x^3} at x = 10

S''(10) > 0

Therefore, Surface is minimum at x = 10 inches

Minimum Surface area, S = 300 in²

7 0
3 years ago
Temperature transducers of a certain type are shipped in batches of 50. A sample of 58 batches was selected, and the number of t
andrew11 [14]

Answer:

a) X        Freq.         Rel Freq.

__________________________

0           7              (7/58) = 0.121

1            10            (10/58)=0.172

2           13            (13/58) =0.224

3           14            (14/58)=0.241

4           6              (6/58)=0.103

5           3              (3/58)=0.052

6           3              (3/58)=0.052

7            1              (1/58) =0.017

8            1              (1/58)=0.017

_________________________

Total     58                 1.00

b) \frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862

Step-by-step explanation:

Assuming this question: Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data:

2,1,2,4,0,1,3,2,0,5,3,3,1,3,2,4,7,0,2,3,

0,4,2,1,3,1,1,3,4,2,3,2,2,8,4,5,1,3,1,

5,0,2,3,2,1,0,6,4,2,1,6,0,3,3,3,6,2,3

(a) Determine frequencies and relative frequencies for the observed values of x = number of nonconforming transducers in a batch. (Round your relative frequencies to three decimal places.)

For this case first we order the dataset on increasing way and we got:

0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1

2 2 2 2  2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3  3 3

4 4 4 4 4 4

5 5 5

6 6 6

7

8

And we can construct the following table:

X        Freq.         Rel Freq.

__________________________

0           7              (7/58) = 0.121

1            10            (10/58)=0.172

2           13            (13/58) =0.224

3           14            (14/58)=0.241

4           6              (6/58)=0.103

5           3              (3/58)=0.052

6           3              (3/58)=0.052

7            1              (1/58) =0.017

8            1              (1/58)=0.017

_________________________

Total     58                 1.00

(b) What proportion of batches in the sample have at most four nonconforming transducers? (Round your answer to three decimal places.)

For this case this proportion would be:

\frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862

4 0
3 years ago
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