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Tasya [4]
3 years ago
15

Which statements represent the rules of significant figures? Check all that apply.

Mathematics
1 answer:
maksim [4K]3 years ago
8 0

Answer:

The integers 1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant.

Zeros in between numbers are always significant.

Step-by-step explanation:

There are three rules by which the number of significant figures can be determined:

  • The digits which are other than the zero are said to be significant figures.
  • Any zero which is present between two significant figure is said to be significant.
  • The zero which appears in the right of the decimal portion in the end is significant. The zero which comes after the decimal is not significant.
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The following dot plot represents student scores on the Unit 1 math test. Scores are in whole number increments.
ratelena [41]

Answer:

38.3 (idk if it can be a decimal)

i think its 15. dont come at me

18 i think

~s9154499~

~Mia for short~

5 0
3 years ago
For which value of m 4/16=m/24
vampirchik [111]
Solve for m:
1/4 = m/24
1/4 = m/24 is equivalent to m/24 = 1/4:
m/24 = 1/4
Multiply both sides of m/24 = 1/4 by 24:
(24 m)/24 = 24/4
(24 m)/24 = 24/24×m = m:
m = 24/4
24/4 = (4×6)/4 = 6:
Answer:  m = 6
5 0
3 years ago
Read 2 more answers
Can someone help me please ?
wlad13 [49]

Answer: dnwkdnewkofbjlsdfsmf

Step-by-step explanation:

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3 years ago
Please help THIS SHOULD BE EASY
goldfiish [28.3K]

Answer:

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Step-by-step explanation:

I hope these r the answers.

4 0
3 years ago
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Consider a series system composed of 4 separate components where each component has a 30% chance of failing. Assume each compone
Marina86 [1]

Answer:

16.15% probability that exactly 3 of them would function

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability of each system working:

4 components, which means that n = 4

Each has a 30% probability of failing, so p = 1 - 0.3 = 0.7

For the system to work, all 4 components have to work. This is P(X = 4).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401

0.2401 probability of a system working.

If you have 7 of these systems, what is the probability that exactly 3 of them would function?

Now 7 systems, so n = 7

0.2401 probability of a system working.

We have to find P(X = 3).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{7,3}.(0.2401)^{3}.(0.7599)^{4} = 0.1615

16.15% probability that exactly 3 of them would function

5 0
3 years ago
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