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pshichka [43]
3 years ago
7

How many seconds are in 2:53

Mathematics
1 answer:
Alik [6]3 years ago
6 0

Answer:

Answer is 10,380

Step-by-step explanation:

If it's 2h53min, then there are 10,380 seconds.

You might be interested in
m∠AOB = 6x + 5, m∠BOC = 4x - 2, m∠AOC = 8x + 21
AnnZ [28]
The sum of angle AOB and angle BOC is equal tot he angle of AOC so

6x+5 +4x-2 = 8x+21

10x+3=8x+21

Subtract three and 8x from both sides

2x=18

Divide by two

x=9

Hope this helped!
6 0
3 years ago
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=Solve%3A%202x%20%2B%207%20%3D%20-11PLS%20HELP%20ILL%20GIVE%20YOU%20BRAINLIEST" id="TexFormula1
Lena [83]

Answer:

-9

Step-by-step explanation:

2x + 7 = -11

2x = -11 - 7

2x = -18

x = -9

7 0
3 years ago
How would I organize the calculation
Viefleur [7K]

Answer:

52 should be the answer

Step-by-step explanation:

first, love the parathesis. subtract 7-2

second 3 raised to the second power

third times 3 raised to the third power by 5 the answer a few u substrates 7-2

fourth 14 divide 2

finally I would add (14/2) 7 + 45 (which is 9 times 5)

read through it carefully by the end if the problem, you should have 52 as the ans

6 0
2 years ago
URGENT PLEASE HELP !!!
VikaD [51]

Answer:

C

Step-by-step explanation:

if you input x into both equations you will get that answer

8 0
3 years ago
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